Motion Tough Numerical Questions Topic Wise Class 9, Test Yourself!

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Motion Tough Numerical Questions Topic Wise Class 9, are designed to test your analytical and conceptual clarity, along with their answers and detailed explanations, all within the syllabus of your NCERT textbook chapter 7 Motion .

Before you solve “Motion Tough Numerical Questions Topic Wise Class 9, Test Yourself!”, ensure that you are thorough in your preparations.

You can check our relevant posts on motion below:

Table of Contents

Distance and Displacement

These questions test the understanding of scalar (distance) vs. vector (displacement) quantities, especially in non-trivial paths.

  1. Question 1: A robotic explorer is sent to survey a flat Martian surface. It starts from its base camp (Point O) and travels 15 km North, then 20 km East, and finally 15 km South.

    • (a) What is the total distance covered by the explorer?

    • (b) What is the magnitude of the explorer’s displacement from its base camp?

    • (c) If the explorer then travels 20 km West from its current position, what will be its final displacement and total distance covered from the initial base camp?

    Answer and Explanation:
    • Conceptual Clarity: This question requires visualizing the path and distinguishing between the total path length (distance) and the shortest straight-line path from start to end (displacement).
    • (a) Total distance covered:
      • Distance North = 15 km
      • Distance East = 20 km
      • Distance South = 15 km
      • Total Distance = 15 km + 20 km + 15 km = 50 km.
    • (b) Magnitude of displacement:
      • The explorer moves 15 km North and then 15 km South. These movements cancel each other out in the North-South direction.
      • The net movement is only 20 km East from the starting point.
      • Therefore, the magnitude of the displacement is 20 km (East).
    • (c) After traveling 20 km West:
      • From its position after the first three legs (20 km East of O), the explorer travels 20 km West. This means it returns exactly to its starting point O.
      • Total Distance = 50 km (from parts a & b) + 20 km (West) = 70 km.
      • Since the final position coincides with the initial position (O), the displacement is zero.
  2. Question 2: A particle moves along a straight line. It starts at point A, travels 40 m to point B, then immediately turns around and travels 70 m back to point C (which is past A). From C, it again reverses direction and moves 20 m back towards A to point D.

    (a) Calculate the total distance covered by the particle.
    (b) Determine the magnitude of the final displacement of the particle from its initial position A.Answer and Explanation
    • Conceptual Clarity: This problem involves multiple changes in direction along a straight line, which is crucial for distinguishing between accumulated path length (distance) and net change in position (displacement).
    • (a) Total distance covered:
      • A to B = 40 m
      • B to C = 70 m
      • C to D = 20 m
      • Total Distance = 40 m + 70 m + 20 m = 130 m.
    • (b) Magnitude of final displacement:
      • Initial position: A (let’s consider A as the origin, 0 m)
      • Movement 1: A to B (+40 m)
      • Movement 2: B to C (-70 m from B, so 40 – 70 = -30 m from A). This means C is 30 m behind A.
      • Movement 3: C to D (+20 m from C, so -30 + 20 = -10 m from A). This means D is 10 m behind A.
      • The final position D is 10 m from the initial position A.
      • The magnitude of the final displacement is 10 m. (Direction is opposite to the initial positive direction assumed, but only magnitude is asked).
  3. Question 3: An object starts at the origin (0,0) and moves in a sequence of straight-line paths:
    • First, it moves 6 m along the positive x-axis.Second, it moves 8 m along the positive y-axis.Third, it moves 6 m along the negative x-axis.Calculate the total distance covered and the magnitude of its final displacement from the origin.

    Answer and Explanation:
    • Conceptual Clarity: This problem requires understanding displacement in two dimensions, which can be visualized as forming parts of a right-angled triangle.
    • Total distance covered:
      • Distance 1 = 6 m
      • Distance 2 = 8 m
      • Distance 3 = 6 m
      • Total Distance = 6 m + 8 m + 6 m = 20 m.
    • Magnitude of final displacement:
      • Path 1: (0,0) to (6,0)
      • Path 2: (6,0) to (6,8)
      • Path 3: (6,8) to (0,8) (since 6m positive x and 6m negative x cancel out).
      • The final position is (0,8).
      • The initial position is (0,0).
      • The shortest distance from (0,0) to (0,8) is 8 m along the positive y-axis.
      • The magnitude of the final displacement is 8 m.
  4. Question 4: An athlete jogs 3/4ths of the way around a circular track of diameter 100 m.

    (a) What is the total distance covered by the athlete?(b) What is the magnitude of the athlete’s displacement from their starting point? (Hint: A circular track implies constant radius, and diameter is twice the radius).Answer and Explanation:
  5. Question 5: A bird flies directly from point A to point B, covering a distance of 150 m. Then it rests for 1 minute. After resting, it flies directly from point B to point C, which is 100 m in a direction perpendicular to its first flight path.

    (a) What is the total distance covered by the bird?
    (b) What is the magnitude of the bird’s displacement from its initial position A?
    (c) Does the 1-minute rest period affect the total distance or displacement? Explain why or why not.

    Answer and Explanation:
    • Conceptual Clarity: This question tests understanding that time elapsed affects distance and speed/velocity calculations, but not the final displacement if the start and end points are fixed. It also involves perpendicular paths.
    • (a) Total distance covered:
      • Distance A to B = 150 m
      • Distance B to C = 100 m
      • Total Distance = 150 m + 100 m = 250 m.
    • (b) Magnitude of displacement:
      • The two paths are perpendicular, forming a right-angled triangle.
      • The displacement is the hypotenuse from A to C.
      • Displacement = sqrt((150 m)² + (100 m)²) = sqrt(22500 + 10000) = sqrt(32500) ≈ 180.28 m.
    • (c) Effect of rest period:
      • The 1-minute rest period does not affect the total distance covered because distance is the total path length, regardless of time taken or pauses.
      • The 1-minute rest period does not affect the final displacement because displacement only depends on the initial and final positions, not the intermediate time or pauses.
      • However, it would affect the average speed as the total time taken for the journey would increase.

Speed and Velocity

2. Speed and Velocity

These questions require careful distinction between speed (scalar) and velocity (vector), and their average calculations, often involving multiple segments of motion and unit conversions.

  1. Question 1: A train travels from Station P to Station Q, a distance of 300 km. For the first 100 km, it maintains an average speed of 50 km/h. For the remaining 200 km, due to track work, its average speed drops to 40 km/h.

    (a) What is the total time taken for the entire journey?(b) Calculate the average speed of the train for the entire journey from P to Q.(c) Can you determine the average velocity for the entire journey without more information? Explain.

    Answer and Explanation:
    • Conceptual Clarity: This problem requires calculating the time for each segment and then applying the average speed formula correctly (total distance divided by total time), which is not simply the average of the two speeds. It also highlights the difference between speed and velocity.
  2. (a) Total time taken:
    • Time for first 100 km (t1) = Distance / Speed = 100 km / 50 km/h = 2 h.
    • Time for remaining 200 km (t2) = Distance / Speed = 200 km / 40 km/h = 5 h.
    • Total time = t1 + t2 = 2 h + 5 h = 7 h.
  3. (b) Average speed for the entire journey:
    • Total Distance = 100 km + 200 km = 300 km.
    • Total Time = 7 h (from part a).
    • Average Speed = Total Distance / Total Time = 300 km / 7 h ≈ 42.86 km/h.
  4. (c) Average velocity:
    • No, we cannot determine the average velocity for the entire journey from P to Q with the given information alone.
    • Velocity requires both magnitude (speed) and direction. While we know the train travels from P to Q, which implies a straight line, the problem does not specify that this is the shortest path or if the train returned. Assuming it’s a straight line and no return, the displacement would be 300 km.

      However, the question asks if we can determine it, and without explicit mention of it being a straight path for displacement calculation, it’s ambiguous. Average velocity is displacement/total time. If P to Q is a straight path, then displacement = 300 km, and average velocity = 300 km / 7 h = 42.86 km/h in the direction from P to Q.

      But the ambiguity makes it “tough.” The source defines velocity as speed in a “definite direction”.
  5. Question 2: A swimmer completes a 100 m race. She swims the first 50 m in 25 s. Due to fatigue, she swims the next 50 m in 30 s.
    • (a) Calculate her average speed for the entire race.

    • (b) If the race was in a straight line from one end of the pool to the other, what is her average velocity for the race?

    • (c) If she then immediately swims back to her starting point in 45 s, what is her average speed and average velocity for the entire round trip (100 m out, 100 m back)?

    Answer and Explanation:
    • Conceptual Clarity: This question differentiates average speed and average velocity, particularly for a round trip where displacement can be zero.
    • (a) Average speed for the entire race (100 m):
      • Total Distance = 50 m + 50 m = 100 m.
      • Total Time = 25 s + 30 s = 55 s.
      • Average Speed = Total Distance / Total Time = 100 m / 55 s ≈ 1.82 m/s.
    • (b) Average velocity for the race (straight line 100 m):
      • Displacement = 100 m (since it’s a straight line from start to finish).
      • Total Time = 55 s.
      • Average Velocity = Displacement / Total Time = 100 m / 55 s ≈ 1.82 m/s in the direction of the race. (Note: Magnitude is equal to average speed because motion is along a straight line and in one direction).
    • (c) Average speed and average velocity for the entire round trip:
      • Total Distance for round trip = 100 m (out) + 100 m (back) = 200 m.
      • Total Time for round trip = 55 s (out) + 45 s (back) = 100 s.
      • Average Speed (round trip) = Total Distance / Total Time = 200 m / 100 s = 2 m/s.
      • Displacement for round trip = 0 m (because the swimmer returns to the starting point, so final position coincides with initial position).
      • Average Velocity (round trip) = Displacement / Total Time = 0 m / 100 s = 0 m/s.
  6. Question 3: An automobile’s odometer reads 1500 km at 9:00 am. At 1:00 pm on the same day, its odometer reads 1800 km.
    • (a) Calculate the average speed of the car in km/h.

    • (b) Convert this average speed to m/s.

    • (c) The car then travels an additional 100 km North in 1.5 hours. What is the car’s average velocity for this specific additional segment of the journey in m/s?

    Answer and Explanation:
    • Conceptual Clarity: This problem combines average speed, unit conversions, and then introduces velocity for a specific segment, highlighting the directional aspect.
    • (a) Average speed in km/h:
      • Distance travelled = Final reading – Initial reading = 1800 km – 1500 km = 300 km.
      • Time elapsed = 1:00 pm – 9:00 am = 4 hours.
      • Average Speed = Distance / Time = 300 km / 4 h = 75 km/h.
    • (b) Convert to m/s:
      • 1 km = 1000 m; 1 h = 3600 s.
      • 75 km/h = 75 × (1000 m / 3600 s) = 75 × (5/18) m/s ≈ 20.83 m/s.
    • (c) Average velocity for the additional segment:
      • Distance (and displacement, assuming straight line) = 100 km = 100,000 m.
      • Time = 1.5 hours = 1.5 × 3600 s = 5400 s.
      • Average Velocity = Displacement / Time = 100,000 m / 5400 s ≈ 18.52 m/s North.
  7. Question 4: A bus travels from city A to city B, a distance of 400 km, in 5 hours. It returns from city B to city A in 4 hours, taking a slightly different route that is 420 km long.

    (a) Calculate the average speed of the bus for the entire round trip.
    (b) What is the average velocity of the bus for the entire round trip?

    Answer and Explanation:
    • Conceptual Clarity: This problem again emphasizes the difference between average speed and average velocity for a round trip, with differing path lengths for outbound and return journeys.
    • (a) Average speed for the entire round trip:
      • Total Distance = Distance (A to B) + Distance (B to A) = 400 km + 420 km = 820 km.
      • Total Time = Time (A to B) + Time (B to A) = 5 h + 4 h = 9 h.
      • Average Speed = Total Distance / Total Time = 820 km / 9 h ≈ 91.11 km/h.
    • (b) Average velocity for the entire round trip:
      • Since the bus starts at city A and returns to city A, its final position coincides with its initial position.
      • Therefore, the total displacement for the entire round trip is 0 km.
      • Average Velocity = Total Displacement / Total Time = 0 km / 9 h = 0 km/h.
  8. Question 5: An object moves along a straight line. For the first 10 seconds, it travels at a constant speed of 5 m/s. For the next 15 seconds, it travels at a constant speed of 8 m/s in the same direction.

    (a) What is the total distance covered by the object?
    (b) What is the object’s average speed over the entire 25-second period?
    (c) What is the object’s average velocity over the entire 25-second period?

    Answer and Explanation:
    • Conceptual Clarity: This problem assesses average speed and average velocity for multi-stage motion where direction is consistent, allowing for a direct comparison of their magnitudes.
    • (a) Total distance covered:
      • Distance 1 = Speed × Time = 5 m/s × 10 s = 50 m.
      • Distance 2 = Speed × Time = 8 m/s × 15 s = 120 m.
      • Total Distance = 50 m + 120 m = 170 m.
    • (b) Average speed over the entire period:
      • Total Time = 10 s + 15 s = 25 s.
      • Average Speed = Total Distance / Total Time = 170 m / 25 s = 6.8 m/s.
    • (c) Average velocity over the entire period:
      • Since the motion is along a straight line and in the same direction, the magnitude of the displacement is equal to the total distance covered.
      • Displacement = 170 m.
      • Average Velocity = Displacement / Total Time = 170 m / 25 s = 6.8 m/s in the original direction of motion. In this case, the average speed and average velocity magnitudes are equal because the object does not change direction and moves along a straight line.

Acceleration

These questions focus on the rate of change of velocity, including uniform and non-uniform scenarios, and calculations involving unit conversions.

  1. Question 1: A car accelerates uniformly from 36 km/h to 90 km/h in 10 seconds. After reaching 90 km/h, it immediately begins to decelerate uniformly, coming to a complete stop in an additional 20 seconds.

    (a) Calculate the acceleration of the car in the first 10 seconds in m/s²
    (b) Calculate the acceleration (deceleration) of the car in the subsequent 20 seconds in m/s².
    (c) What is the magnitude of the average acceleration over the entire journey from start to stop?

    Answer and Explanation:
  2. Question 2: A train starts from rest at a station and accelerates uniformly. It reaches a speed of 120 km/h in 5 minutes.

    (a) Calculate the acceleration of the train in m/s².
    (b) If the train then travels at this constant speed for 15 minutes, what is its acceleration during this period?
    (c) If, after this constant speed period, the train applies brakes and decelerates at a rate of -0.5 m/s², how long will it take to come to a complete stop?

    Answer and Explanation:
    • Conceptual Clarity: This problem combines different phases of motion (acceleration, constant velocity, deceleration) and requires consistent unit handling and application of acceleration definition.
    • Initial conversions:
      • 120 km/h = 120 × (1000/3600) m/s = 100/3 m/s ≈ 33.33 m/s.
      • 5 minutes = 5 × 60 s = 300 s.
      • 15 minutes = 15 × 60 s = 900 s.
    • (a) Acceleration in the first 5 minutes:
      • Initial velocity (u) = 0 m/s (starts from rest).
      • Final velocity (v) = 100/3 m/s.
      • Time (t) = 300 s.
      • Acceleration (a) = (v – u) / t = (100/3 m/s – 0 m/s) / 300 s = (100/3) / 300 m/s² = 100 / 900 m/s² = 1/9 m/s² (approx. 0.11 m/s²).
    • (b) Acceleration during constant speed travel:
      • During uniform motion, velocity remains constant.
      • If velocity is constant, the change in velocity is zero.
      • Therefore, the acceleration is 0 m/s².
    • (c) Time to stop during deceleration:
      • Initial velocity (u) = 100/3 m/s (speed before braking)
      • Final velocity (v) = 0 m/s (comes to a complete stop)
      • Acceleration (a) = -0.5 m/s²
      • Using a = (v – u) / t, we can rearrange to find t = (v – u) / a.
      • t = (0 m/s – 100/3 m/s) / (-0.5 m/s²) = (-100/3) / (-1/2) s = (100/3) × 2 s = 200/3 s ≈ 66.67 seconds.
  3. Question 3: An object’s velocity changes from 20 m/s East to 20 m/s South in 5 seconds.

    (a) What is the magnitude of the object’s initial speed and final speed?
    (b) Is this motion uniform or non-uniform? Explain.
    (c) Calculate the magnitude and direction of the average acceleration during this 5-second interval.
    (Hint: Acceleration is a vector, consider change in velocity vectorially).

    Answer and Explanation:
    • Conceptual Clarity: This question is tough because it deals with acceleration due to a change in direction of velocity, even if the speed (magnitude) remains constant. This requires vector subtraction for velocity change.
    • (a) Initial and final speed:
      • Speed is the magnitude of velocity.
      • Initial speed = 20 m/s.
      • Final speed = 20 m/s.
    • (b) Uniform or non-uniform motion:
      • This motion is non-uniform. Although the speed remains constant, the velocity changes because its direction changes (from East to South). Therefore, the object is undergoing acceleration.
    • (c) Magnitude and direction of average acceleration:
      • Initial velocity (u) = 20 m/s East (let’s represent as (20, 0) if East is +x and North is +y)
      • Final velocity (v) = 20 m/s South (let’s represent as (0, -20))
      • Change in velocity (Δv) = v – u = (0, -20) – (20, 0) = (-20, -20) m/s.
      • Magnitude of change in velocity = sqrt((-20)² + (-20)²) = sqrt(400 + 400) = sqrt(800) ≈ 28.28 m/s.
      • Time (t) = 5 s.
      • Magnitude of Average Acceleration = Magnitude of (Δv / t) = 28.28 m/s / 5 s ≈ 5.66 m/s². Ans
      • Direction of change in velocity: Since Δv is (-20, -20), it points South-West (45° South of West). The acceleration is in this same direction.
  4. Question 4: A stone is thrown vertically upward with an initial velocity of 15 m/s. If the acceleration due to gravity is 10 m/s² in the downward direction.

    (a) What will be the velocity of the stone after 1.0 second?
    (b) How much time will it take for the stone to reach its highest point (where its velocity becomes zero)?
    (c) What will be its velocity just before it hits the ground if it returns to its initial throwing height?Answer and Explanation:
    • Conceptual Clarity: This problem directly tests the understanding of uniform acceleration against the direction of motion, as described for a freely falling body, specifically for vertical motion where acceleration is constant and downwards.
    • Assumptions: Upward direction is positive, so downward acceleration is negative. a = -10 m/s².
    • (a) Velocity after 1.0 second:
      • Initial velocity (u) = 15 m/s
      • Acceleration (a) = -10 m/s²
      • Time (t) = 1.0 s
      • Using v = u + at:
      • v = 15 m/s + (-10 m/s²) × 1.0 s = 15 m/s – 10 m/s = 5 m/s (upwards). Ans
    • (b) Time to reach highest point:
      • Initial velocity (u) = 15 m/s
      • Final velocity (v) = 0 m/s (at highest point)
      • Acceleration (a) = -10 m/s²
      • Using v = u + at, we find t = (v – u) / a.
      • t = (0 m/s – 15 m/s) / (-10 m/s²) = -15 / -10 s = 1.5 seconds. Ans
    • (c) Velocity just before hitting the ground:
      • The motion is symmetrical. The time taken to go up to the highest point is 1.5 s. The time taken to fall back from the highest point to the initial height is also 1.5 s. So, total time in air = 3.0 s.
      • Initial velocity for the entire trip (u) = 15 m/s
      • Acceleration (a) = -10 m/s²
      • Total time (t) = 3.0 s
      • Using v = u + at:
      • v = 15 m/s + (-10 m/s²) × 3.0 s = 15 m/s – 30 m/s = -15 m/s.
      • The negative sign indicates the velocity is in the downward direction. The speed is 15 m/s, same as initial, but direction is opposite.
  5. Question 5: A vehicle starts from rest and moves with uniform acceleration. In the first 5 seconds, it covers a distance of 25 meters.

    (a) Calculate the acceleration of the vehicle.
    (b) What will be its velocity at the end of these 5 seconds?
    (c) If the acceleration then becomes non-uniform and its velocity changes by unequal amounts in equal intervals of time, what type of motion is this?

Answer and Explanation:

    ◦ Conceptual Clarity: This question links distance, time, initial velocity, and acceleration, requiring the use of the appropriate equation of motion. It also tests the definition of non-uniform acceleration.

    ◦ (a) Acceleration of the vehicle:

        Initial velocity (u) = 0 m/s (starts from rest).

Distance (s) = 25 m.

        Time (t) = 5 s.

        Using the equation s = ut + ½ at²:

        25 m = (0 m/s)(5 s) + ½ a (5 s)²

        25 m = 0 + ½ a (25 s²)

        25 m = 12.5 a s²

        a = 25 m / 12.5 s² = 2 m/s². Ans

    ◦ (b) Velocity at the end of 5 seconds:

        ▪ Initial velocity (u) = 0 m/s.

        ▪ Acceleration (a) = 2 m/s² (from part a).

        ▪ Time (t) = 5 s.

        ▪ Using the equation v = u + at:

        ▪ v = 0 m/s + (2 m/s²)(5 s) = 10 m/s.

    ◦ (c) Type of motion if acceleration becomes non-uniform:

        ▪ If the velocity changes at a non-uniform rate, the object is said to be moving with non-uniform acceleration.

Graphical Representation of Motion

These questions require interpreting information from distance-time and velocity-time graphs, and calculating quantities from them.

  1. Question 1: Consider a Distance-Time graph for an object. The graph has three segments:
    • Segment 1: From t=0 s to t=5 s, the distance increases linearly from 0 m to 10 m.Segment 2: From t=5 s to t=10 s, the distance remains constant at 10 m.Segment 3: From t=10 s to t=15 s, the distance increases linearly from 10 m to 35 m.

    • (a) Describe the motion of the object in each segment (uniform speed, at rest, non-uniform speed).

    • (b) Calculate the speed of the object in Segment 1 and Segment 3.

    • (c) Calculate the average speed of the object for the entire 15-second journey.
    Answer and Explanation:
    • Conceptual Clarity: This problem requires interpreting different slopes on a distance-time graph, including zero slope for rest and calculating speed from slope.
    • (a) Description of motion in each segment:
      • Segment 1 (t=0 to t=5 s): The graph is a straight line with a positive slope, indicating uniform speed.
      • Segment 2 (t=5 to t=10 s): The graph is a horizontal straight line, meaning the distance does not change with time. This indicates the object is at rest.
      • Segment 3 (t=10 to t=15 s): The graph is a straight line with a steeper positive slope than Segment 1. This indicates uniform speed. (Note: If it were curved, it would be non-uniform).
    • (b) Speed of the object:
      • Segment 1: Speed (v) = (Change in distance) / (Change in time) = (10 m – 0 m) / (5 s – 0 s) = 10 m / 5 s = 2 m/s.
      • Segment 3: Speed (v) = (Change in distance) / (Change in time) = (35 m – 10 m) / (15 s – 10 s) = 25 m / 5 s = 5 m/s.
    • (c) Average speed for the entire journey:
      • Total Distance Covered = 35 m (final distance from origin).
      • Total Time Taken = 15 s.
      • Average Speed = Total Distance / Total Time = 35 m / 15 s ≈ 2.33 m/s.
  2. Question 2: A Velocity-Time graph shows the following motion for a car:
    • Segment 1: From t=0 s to t=10 s, velocity increases linearly from 0 m/s to 20 m/s.Segment 2: From t=10 s to t=20 s, velocity remains constant at 20 m/s.Segment 3: From t=20 s to t=25 s, velocity decreases linearly from 20 m/s to 0 m/s.

    • (a) Calculate the acceleration of the car in each segment.(b) Calculate the total displacement of the car during the entire 25-second journey by finding the area under the graph.(c) What is the average velocity of the car for the entire journey?

    Answer and Explanation:
    • Conceptual Clarity: This problem is a comprehensive test of velocity-time graph interpretation, including calculating acceleration from slope and displacement from area, and then deriving average velocity.
    • (a) Acceleration in each segment:
      • Segment 1 (t=0 to t=10 s): Acceleration (a) = (Change in velocity) / (Change in time) = (20 m/s – 0 m/s) / (10 s – 0 s) = 20 m/s / 10 s = 2 m/s².
      • Segment 2 (t=10 to t=20 s): Velocity is constant, so acceleration = 0 m/s².
      • Segment 3 (t=20 to t=25 s): Acceleration (a) = (Change in velocity) / (Change in time) = (0 m/s – 20 m/s) / (25 s – 20 s) = -20 m/s / 5 s = -4 m/s². (Negative indicates deceleration).
    • (b) Total displacement (area under graph):
      • Segment 1 (Triangle): Area = ½ × Base × Height = ½ × 10 s × 20 m/s = 100 m.
      • Segment 2 (Rectangle): Area = Base × Height = (20 s – 10 s) × 20 m/s = 10 s × 20 m/s = 200 m.
      • Segment 3 (Triangle): Area = ½ × Base × Height = ½ × (25 s – 20 s) × 20 m/s = ½ × 5 s × 20 m/s = 50 m.
      • Total Displacement = 100 m + 200 m + 50 m = 350 m.
    • (c) Average velocity for the entire journey:
      • Average Velocity = Total Displacement / Total Time.
      • Total Time = 25 s.
      • Average Velocity = 350 m / 25 s = 14 m/s.
  3. Question 3: Refer to Fig. 7.10 (Distance-time graph for three objects A, B, and C).

    (a) Rank the three objects (A, B, C) from fastest to slowest.
    (b) How much distance has object A covered when object B passes object C?
    (c) At what approximate time does object B overtake object A?

    Answer and Explanation:
    • Conceptual Clarity: This question requires careful visual interpretation of slopes on a distance-time graph to compare speeds and find points of intersection (overtaking).
    • (a) Ranking from fastest to slowest:
      • Speed from a distance-time graph is determined by the slope (steeper slope = higher speed).
      • Visually, the slope of B is the steepest, followed by A, and then C.
      • Therefore, the ranking is: B > A > C (B is fastest, C is slowest).
    • (b) Distance covered by A when B passes C:
      • Locate the point where lines B and C intersect. This is approximately at a distance of 4 units (km or m, units not specified in fig but common in text).
      • At this point, follow the vertical line downwards to the time axis, which is approximately at t=4.5 units (hours or seconds).
      • Now, locate object A’s position at this time (t=4.5 units). Follow the vertical line up from t=4.5 to line A.
      • Then follow the horizontal line from that point to the distance axis. Object A is approximately at a distance of 7 units.
    • (c) Approximate time B overtakes A:
      • Locate the intersection point of lines A and B.
      • Follow the vertical line downwards from this intersection point to the time axis.
      • The approximate time is around t=5.7 to 5.8 units.
  4. Question 4: A car travels along a straight road. It starts from rest and maintains a uniform acceleration of 2 m/s² for 5 seconds. Then, it travels at a constant velocity for the next 10 seconds. Finally, it decelerates uniformly at 4 m/s² until it comes to rest.

    (a) Sketch a velocity-time graph for this motion.
    (b) Using your graph, determine the maximum velocity attained by the car.
    (c) Using your graph, calculate the total time taken for the entire journey.

    Answer and Explanation:
    • Conceptual Clarity: This problem requires translating a verbal description of motion into a velocity-time graph and then extracting key numerical information from the graph.
    • (a) Sketch of the velocity-time graph:
      • Segment 1 (Acceleration): From t=0, v=0. For 5s, a=2 m/s². Final velocity v = u + at = 0 + (2)(5) = 10 m/s. So, a straight line from (0,0) to (5s, 10m/s).
      • Segment 2 (Constant Velocity): From t=5s, v=10m/s. For next 10s, velocity is constant. So, horizontal line from (5s, 10m/s) to (15s, 10m/s).
      • Segment 3 (Deceleration): From t=15s, v=10m/s. Decelerates at -4 m/s² until v=0. Time taken for deceleration: t = (v – u) / a = (0 – 10) / (-4) = 2.5 s. So, a straight line from (15s, 10m/s) to (17.5s, 0m/s).
      • (Self-Correction: The sketch cannot be provided as an image, but the description forms the basis for visualization and calculation.)
    • (b) Maximum velocity attained:
      • The maximum velocity is reached at the end of the acceleration phase and maintained during the constant velocity phase.
      • From Segment 1 calculation, the velocity reaches 10 m/s. This is the peak of the graph before it becomes horizontal.
    • (c) Total time taken for the entire journey:
      • Time for acceleration = 5 s.
      • Time for constant velocity = 10 s.
      • Time for deceleration = 2.5 s (calculated above).
      • Total Time = 5 s + 10 s + 2.5 s = 17.5 seconds.
  5. Question 5: A distance-time graph for a runner is shown below (imagine a graph where the distance increases slowly at first, then more rapidly, and then slowly again, forming an ‘S’ like curve, suggesting non-uniform motion).

    (a) Describe the type of motion represented by this graph.
    (b) If the graph is not a straight line, can the speed be directly read off the graph at any instant? Explain how average speed over an interval would be found.
    (c) If the graph passes through points (2s, 5m) and (6s, 45m), what is the average speed between these two points?
    • Answer and Explanation:
      • Conceptual Clarity: This question probes the understanding of non-linear distance-time graphs, the difference between instantaneous and average speed, and how to calculate average speed from such a graph.
      • (a) Type of motion:
        • Since the graph is a curve (non-linear variation of distance with time), it represents non-uniform speed (or non-uniform motion).
      • (b) Reading instantaneous speed and finding average speed:
        • No, the speed cannot be directly read off the graph at any instant if it’s a curved line. The slope of the tangent at any point would represent the instantaneous speed, but the source doesn’t detail tangents.
        • To find the average speed over an interval, one would calculate the total distance covered during that interval and divide by the total time taken for that interval. This is the slope of the straight line connecting the start and end points of the interval on the distance-time graph.
      • (c) Average speed between (2s, 5m) and (6s, 45m):
        • Distance covered (s₂ – s₁) = 45 m – 5 m = 40 m.
        • Time taken (t₂ – t₁) = 6 s – 2 s = 4 s.
        • Average Speed (v) = (s₂ – s₁) / (t₂ – t₁) = 40 m / 4 s = 10 m/s.

Equations of Motion

These questions require selecting the appropriate equation of motion and performing calculations, often involving multiple steps or unit conversions.

  1. Question 1: A car applies brakes and produces a uniform acceleration of -5 m/s². If the car was travelling at 72 km/h when the brakes were applied, calculate:
    • (a) The time it takes for the car to come to a complete stop.
    • (b) The distance the car travels during this braking period.
    • (c) If a different car, starting from rest, covers the same distance calculated in (b) in 4 seconds, what would be its uniform acceleration?
    Answer and Explanation:
    • Conceptual Clarity: This problem involves deceleration, calculation of stopping time and distance using equations of motion, and then applying those values to a new scenario with different initial conditions.
    • Initial conversion:
      • 72 km/h = 72 × (1000/3600) m/s = 20 m/s.
    • (a) Time to stop:
      • Initial velocity (u) = 20 m/s.
      • Final velocity (v) = 0 m/s (comes to a stop).
      • Acceleration (a) = -5 m/s².
      • Using v = u + at:
      • 0 = 20 + (-5)t
      • 5t = 20
      • t = 4 seconds.
    • (b) Distance travelled during braking:
      • Using 2as = v² – u²:
      • 2(-5)s = (0)² – (20)²
      • -10s = -400
      • s = 40 meters.
      • Alternatively, using s = ut + ½ at² and t=4s from (a):
      • s = (20)(4) + ½(-5)(4)² = 80 + ½(-5)(16) = 80 – 40 = 40 meters.
    • (c) Acceleration of the different car:
      • Initial velocity (u) = 0 m/s (starts from rest).
      • Distance (s) = 40 m (from part b).
      • Time (t) = 4 s.
      • Using s = ut + ½ at²:
      • 40 = (0)(4) + ½ a (4)²
      • 40 = ½ a (16)
      • 40 = 8a
      • a = 40 / 8 = 5 m/s².
  2. Question 2: A train starts from rest at Station X and moves with a uniform acceleration of 0.2 m/s² for 3 minutes. It then maintains this constant speed for 10 minutes. Finally, it applies brakes and decelerates uniformly at -0.4 m/s² until it comes to a complete stop at Station Y.
    • (a) Calculate the maximum speed attained by the train.
    • (b) Calculate the total distance covered by the train from Station X to Station Y.
    • (c) What is the total time taken for the entire journey from X to Y?
    Answer and Explanation:
    • Conceptual Clarity: This is a multi-stage problem requiring calculation for each phase of motion (acceleration, constant speed, deceleration) using appropriate equations and summing up distances and times.
    • Initial conversions:
      • 3 minutes = 3 × 60 s = 180 s.
      • 10 minutes = 10 × 60 s = 600 s.
    • (a) Maximum speed attained (end of acceleration phase):
      • Initial velocity (u) = 0 m/s.
      • Acceleration (a) = 0.2 m/s².
      • Time (t) = 180 s.
      • Using v = u + at:
      • v = 0 + (0.2)(180) = 36 m/s. This is the maximum speed.
    • (b) Total distance covered:
      • Phase 1 (Acceleration):
        • s₁ = ut + ½ at² = (0)(180) + ½(0.2)(180)² = 0.1 × 32400 = 3240 m.
        • Alternatively, using 2as = v² – u²: 2(0.2)s₁ = (36)² – (0)² => 0.4s₁ = 1296 => s₁ = 3240 m.
      • Phase 2 (Constant Speed):
        • Speed = 36 m/s.
        • Time = 600 s.
        • s₂ = speed × time = 36 × 600 = 21600 m.
      • Phase 3 (Deceleration):
        • Initial velocity (u) = 36 m/s.
        • Final velocity (v) = 0 m/s.
        • Acceleration (a) = -0.4 m/s².
        • Using 2as = v² – u²:
        • 2(-0.4)s₃ = (0)² – (36)²
        • -0.8s₃ = -1296
        • s₃ = 1296 / 0.8 = 1620 m.
      • Total Distance = s₁ + s₂ + s₃ = 3240 m + 21600 m + 1620 m = 26460 m = 26.46 km.
    • (c) Total time taken for the entire journey:
      • Time for Phase 1 = 180 s.
      • Time for Phase 2 = 600 s.
      • Time for Phase 3 (Deceleration): Using v = u + at => 0 = 36 + (-0.4)t => 0.4t = 36 => t = 36 / 0.4 = 90 s.
      • Total Time = 180 s + 600 s + 90 s = 870 seconds = 14 minutes 30 seconds.
  3. Question 3: An object is dropped from a height of 80 m. Assume acceleration due to gravity is 10 m/s² downwards.
    • (a) What is the velocity of the object just before it strikes the ground?
    • (b) How long does it take for the object to reach the ground?
    • (c) If the object was instead thrown downwards with an initial velocity of 5 m/s, what would be its velocity upon striking the ground?
    Answer and Explanation:
    • Conceptual Clarity: This problem applies equations of motion to free fall, considering initial conditions (dropped vs. thrown) and the effect of uniform acceleration due to gravity.
    • Assumptions: Downward direction is positive.
    • (a) Velocity just before striking the ground (dropped):
      • Initial velocity (u) = 0 m/s (dropped from rest).
      • Distance (s) = 80 m.
      • Acceleration (a) = 10 m/s².
      • Using 2as = v² – u²:
      • 2(10)(80) = v² – (0)²
      • 1600 = v²
      • v = sqrt(1600) = 40 m/s (downwards).
    • (b) Time to reach the ground (dropped):
      • Initial velocity (u) = 0 m/s.
      • Final velocity (v) = 40 m/s (from part a).
      • Acceleration (a) = 10 m/s².
      • Using v = u + at:
      • 40 = 0 + 10t
      • t = 40 / 10 = 4 seconds.
    • (c) Velocity upon striking the ground (thrown downwards):
      • Initial velocity (u) = 5 m/s (thrown downwards).
      • Distance (s) = 80 m.
      • Acceleration (a) = 10 m/s².
      • Using 2as = v² – u²:
      • 2(10)(80) = v² – (5)²
      • 1600 = v² – 25
      • v² = 1625
      • v = sqrt(1625) ≈ 40.31 m/s (downwards).
  4. Question 4: Two cars, A and B, are on a straight road. Car A is initially at rest at the origin. Car B is 100 m ahead of Car A and is moving at a constant speed of 10 m/s. Car A starts accelerating uniformly at 2 m/s² at the same instant Car B passes the origin.
    • (a) Write the position equations for both cars in terms of time (t).
    • (b) How long does it take for Car A to catch up to Car B?
    • (c) How far from the origin do they meet?
    Answer and Explanation:
    • Conceptual Clarity: This problem requires setting up and solving simultaneous equations (or equating position functions) for two objects moving differently to find their meeting point and time.
    • (a) Position equations:
      • Car A:
        • Initial position (s₀) = 0 m.
        • Initial velocity (u) = 0 m/s.
        • Acceleration (a) = 2 m/s².
        • Using s = ut + ½ at²:
        • s_A = 0 + ½ (2)t² = t²
      • Car B:
        • Initial position (s₀) = 100 m (100m ahead of A).
        • Initial velocity (u) = 10 m/s (constant speed means a=0).
        • Using s = ut + ½ at² (or s = s₀ + ut for constant velocity):
        • s_B = 100 + 10t
    • (b) Time for Car A to catch up to Car B:
      • They meet when their positions are equal: s_A = s_B.
      • t² = 100 + 10t
      • t² – 10t – 100 = 0
      • Using the quadratic formula t = [-b ± sqrt(b² – 4ac)] / 2a:
      • t = [10 ± sqrt((-10)² – 4(1)(-100))] / 2(1)
      • t = [10 ± sqrt(100 + 400)] / 2
      • t = [10 ± sqrt(500)] / 2
      • t = [10 ± 22.36] / 2
      • Since time cannot be negative, t = (10 + 22.36) / 2 = 32.36 / 2 = 16.18 seconds.
    • (c) Distance from the origin they meet:
      • Substitute t = 16.18 s into either position equation:
      • Using s_A = t² = (16.18)² ≈ 261.79 meters.
      • (Check with s_B = 100 + 10(16.18) = 100 + 161.8 = 261.8 meters. The values are consistent.)
  5. Question 5: An object travels a distance of 40 m in 8 seconds. Its final velocity is 12 m/s. Assuming uniform acceleration:
    • (a) What was the initial velocity of the object?
    • (b) Calculate the uniform acceleration of the object.
    Answer and Explanation:
    • Conceptual Clarity: This problem requires using two equations of motion simultaneously or strategically to solve for two unknowns (initial velocity and acceleration).
    • (a) Initial velocity of the object:
      • We have s = 40 m, t = 8 s, v = 12 m/s. We need u and a.
      • From v = u + at, we have 12 = u + 8a (Equation 1)
      • From s = ut + ½ at², we have 40 = u(8) + ½ a (8)²
      • 40 = 8u + ½ a (64)
      • 40 = 8u + 32a (Equation 2)
      • From Equation 1, u = 12 – 8a. Substitute this into Equation 2:
      • 40 = 8(12 – 8a) + 32a
      • 40 = 96 – 64a + 32a
      • 40 = 96 – 32a
      • 32a = 96 – 40
      • 32a = 56
      • a = 56 / 32 = 7 / 4 = 1.75 m/s² (We’ll use this in part b).
      • Now substitute ‘a’ back into Equation 1 to find ‘u’:
      • u = 12 – 8(1.75) = 12 – 14 = -2 m/s.
      • The initial velocity was -2 m/s (meaning it was moving in the opposite direction of its final velocity, or decelerating from negative to positive if it passed through zero, or simply starting backwards).
    • (b) Uniform acceleration of the object:
      • From the calculations in part (a), the acceleration (a) is 1.75 m/s².

Uniform Circular Motion

These questions explore the concept of uniform circular motion, where speed is constant but velocity and acceleration change due to continuous change in direction.

  1. Question 1: An athlete runs on a circular track of radius 70 m. They complete one full round in 22 seconds.
    • (a) Calculate the speed of the athlete.
    • (b) What is the magnitude of the athlete’s displacement after completing half a round?
    • (c) Is the athlete’s motion accelerated? Explain.
    Answer and Explanation:
    • Conceptual Clarity: This question combines calculations for speed in circular motion with conceptual understanding of displacement and acceleration in such motion.
    • (a) Speed of the athlete:
      • Radius (r) = 70 m.
      • Time for one round (t) = 22 s.
      • Circumference (distance for one round) = 2πr.
      • Speed (v) = Distance / Time = 2πr / t.
      • v = 2 × (22/7) × 70 m / 22 s = 2 × 22 × 10 m / 22 s = 20 m/s.
    • (b) Magnitude of displacement after half a round:
      • After half a round, the athlete is at the diametrically opposite point from the starting position.
      • The shortest distance between these two points is the diameter of the circle.
      • Diameter = 2 × Radius = 2 × 70 m = 140 m.
    • (c) Is the athlete’s motion accelerated?
      • Yes, the athlete’s motion is accelerated.
      • Although the speed (magnitude of velocity) is constant, the direction of motion is continuously changing at every point along the circular path. Since velocity is a quantity that specifies both speed and direction, a change in direction constitutes a change in velocity, and a change in velocity results in acceleration.
  2. Question 2: An artificial satellite is moving in a circular orbit of radius 42,250 km.
    • (a) Calculate its speed if it completes one revolution in 24 hours. Express the answer in km/h and m/s.
    • (b) Explain why a satellite in uniform circular motion is considered to be accelerating.
    Answer and Explanation:
    • Conceptual Clarity: This problem applies the circular motion speed formula to a real-world example, requiring unit conversions, and tests the conceptual understanding of acceleration in this context.
    • (a) Speed of the satellite:
      • Radius (r) = 42,250 km.
      • Time for one revolution (t) = 24 hours.
      • Speed (v) = 2πr / t.
      • v = 2 × π × 42,250 km / 24 h = 84,500π / 24 km/h ≈ 11,065.4 km/h.
      • To convert to m/s: 1 km = 1000 m, 1 h = 3600 s.
      • 11,065.4 km/h × (1000 m / 1 km) × (1 h / 3600 s) ≈ 3073.7 m/s.
    • (b) Why is it accelerating?
      • Even though the satellite moves with a uniform (constant) speed, its direction of motion is continuously changing as it moves along the circular orbit.
      • Velocity is defined as speed with a definite direction. Since the direction of the satellite’s velocity is constantly changing, its velocity itself is changing.
      • Acceleration is the rate of change of velocity. Therefore, because its velocity is changing (due to direction change), the satellite is constantly accelerating.
  3. Question 3: A string of length 0.8 m has a stone tied to one end. A child holds the other end and swings the stone in a horizontal circle, completing 4 revolutions in 5 seconds.
    • (a) Calculate the speed of the stone.
    • (b) If the child suddenly releases the string at a particular instant, describe the path the stone will take immediately after release.
    • (c) Will the stone’s acceleration be zero immediately after release? Explain.
    Answer and Explanation:
    • Conceptual Clarity: This question combines calculating speed for multiple revolutions with a conceptual understanding of inertia and what happens when uniform circular motion is interrupted.
    • (a) Speed of the stone:
      • Radius (r) = length of the string = 0.8 m.
      • Total revolutions = 4.
      • Total time = 5 s.
      • Time for one revolution (period, t_rev) = Total time / Total revolutions = 5 s / 4 = 1.25 s.
      • Speed (v) = 2πr / t_rev.
      • v = 2 × π × 0.8 m / 1.25 s ≈ 4.02 m/s.
    • (b) Path after release:
      • Immediately after release, the stone will move along a straight line tangent to the circular path at the point where it was released. This is because the stone continues to move along the direction it was moving at that exact instant due to inertia.
    • (c) Acceleration immediately after release:
      • No, the stone’s acceleration will not be zero immediately after release if it is released in a gravitational field (like on Earth).
      • While the circular acceleration (due to change in direction) ceases because the constraint of the string is removed, the stone will now be under the influence of acceleration due to gravity (approximately 10 m/s² downwards, as seen in other problems from the source). Unless it’s in a zero-gravity environment, it will accelerate downwards. If ignoring gravity, and assuming it’s perfectly horizontal and no air resistance, it would move with constant velocity, implying zero acceleration. But the context of “everyday life” suggests gravity is present.
  4. Question 4: Consider an athlete running at a constant speed along different tracks: a square track, a hexagonal track, and a circular track, each having approximately the same perimeter.
    • (a) In which type of track (square, hexagonal, or circular) does the athlete change their direction of motion most frequently?
    • (b) In which type of track (square, hexagonal, or circular) is the athlete undergoing continuous acceleration despite constant speed?
    • (c) What happens to the number of times the athlete changes direction as the number of sides of a polygonal track increases indefinitely?
    Answer and Explanation:
    • Conceptual Clarity: This problem directly references the conceptual development of uniform circular motion from polygons, highlighting the increasing frequency of direction change and continuous acceleration.
    • (a) Most frequent direction change:
      • On a square track, the athlete changes direction 4 times per round.
      • On a hexagonal track, the athlete changes direction 6 times per round.
      • On a circular track, the direction of motion is continuously changing at every point.
      • Therefore, the athlete changes their direction of motion most frequently (continuously) on a circular track.
    • (b) Continuous acceleration:
      • The athlete is undergoing continuous acceleration on the circular track.
      • While moving on straight segments of polygonal tracks, their acceleration would be zero if speed is constant. Acceleration only occurs at the corners where direction changes.
      • However, in uniform circular motion, the direction changes continuously, meaning the velocity is constantly changing, resulting in continuous acceleration.
    • (c) Effect of increasing sides:
      • As the number of sides of the track increases indefinitely, the shape of the track approaches the shape of a circle. Consequently, the athlete has to take turns more and more often, eventually leading to a continuous change in direction (as in a circle).
  5. Question 5: A merry-go-round has a radius of 4 meters. A child is sitting on an animal at the edge of the merry-go-round. The merry-go-round completes 1 revolution every 10 seconds.
    • (a) What is the speed of the child?
    • (b) If the merry-go-round completes 20 revolutions, what is the total distance traveled by the child?
    • (c) What is the total displacement of the child after completing these 20 revolutions, assuming the child remains at the same position relative to the merry-go-round’s center?
    Answer and Explanation:
    • Conceptual Clarity: This question applies circular motion calculations to a common scenario and re-emphasizes the distinction between distance and displacement over multiple complete cycles.
    • (a) Speed of the child:
      • Radius (r) = 4 m.
      • Time for one revolution (t) = 10 s.
      • Speed (v) = 2πr / t.
      • v = 2 × π × 4 m / 10 s = 8π / 10 m/s = 0.8π m/s (approx. 2.51 m/s).
    • (b) Total distance traveled by the child for 20 revolutions:
      • Distance per revolution = Circumference = 2πr = 2 × π × 4 m = 8π m.
      • Total Distance = Number of revolutions × Distance per revolution = 20 × 8π m = 160π m (approx. 502.65 m).
    • (c) Total displacement of the child after 20 revolutions:
      • After each full revolution, the child returns to their initial starting position on the merry-go-round relative to the ground.
      • Since 20 revolutions is a whole number of complete cycles, the final position coincides exactly with the initial position.
      • Therefore, the total displacement of the child is 0 meters.

Conclusion: Motion Tough Numerical Questions Topic Wise Class 9

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