Rate of Change of Velocity — Long Answer Questions (Class 9 Physics)
Medium Level (Application & Explanation)
Q1. Explain the difference between uniform motion and non-uniform motion. Give clear real-life examples and describe how each relates to the concept of acceleration.
Answer:
Uniform motion means an object moves with constant velocity. Its speed and direction do not change, so the change in velocity is zero and thus acceleration is zero. Example: a car cruising steadily at 50 km/h on a straight, empty highway.
Non-uniform motion means the velocity varies with time; speed or direction (or both) change. This produces a non-zero acceleration. Example: a bicycle speeding up from rest to 20 km/h or a car in city traffic that speeds up and slows down.
In uniform motion there is no acceleration; in non-uniform motion acceleration describes the rate at which velocity changes. Understanding which motion applies helps determine whether to compute a = (v − u)/t or analyze changing speeds over time.
Q2. Show how the formula a = (v − u)/t is obtained and explain the SI unit of acceleration. Solve a short numerical example using this formula.
Answer:
The change in velocity over a time interval equals final velocity minus initial velocity, so change in velocity = v − u. Dividing this change by the time taken gives acceleration: a = (v − u)/t.
The SI unit of velocity is m/s, and time is s, so acceleration has unit m/s² (meters per second squared).
Example: A car speeds from 20 m/s to 40 m/s in 5 s. Change in velocity = 40 − 20 = 20 m/s. Using the formula, a = 20 / 5 = 4 m/s².
This shows how to compute the average acceleration when velocity changes uniformly over the time.
Q3. Describe positive acceleration and negative acceleration (deceleration) with everyday examples. Explain why deceleration is not always negative acceleration in vector terms.
Answer:
Positive acceleration occurs when acceleration acts in the same direction as velocity, causing the speed to increase. Example: a car pressing the gas pedal to speed up on a straight road.
Negative acceleration or deceleration commonly means the speed decreases; acceleration acts opposite to velocity. Example: braking to stop at a red light.
In vector terms, acceleration can be positive or negative depending on the chosen direction. If we choose forward as positive, braking gives negative acceleration. But if an object moves backward and slows down, acceleration might be positive while the speed decreases. Thus, “deceleration” describes a decrease in speed, while the sign of acceleration depends on the
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direction selected.
Q4. Explain how a freely falling object shows uniform acceleration. Include the standard value of gravitational acceleration and a short calculation for velocity after a certain time.
Answer:
A freely falling object near Earth (ignoring air resistance) experiences a constant acceleration due to gravity, denoted g, which is approximately 9.8 m/s² downward.
This constant acceleration means the object’s velocity increases by equal amounts in equal time intervals; hence motion is uniformly accelerated.
Example calculation: An object dropped from rest (u = 0). After 3 s, increase in velocity = g × t = 9.8 × 3 = 29.4 m/s downward. Final velocity v ≈ 29.4 m/s downward.
This shows uniform acceleration because each second the velocity increases by about 9.8 m/s, producing predictable motion values.
Q5. What is non-uniform acceleration? Explain how you would find average acceleration for a car whose speed changes irregularly over time.
Answer:
Non-uniform acceleration means the velocity changes by unequal amounts in equal time intervals; acceleration is not constant. Example: a car in city traffic that speeds up, slows down, and idles unpredictably.
To find average acceleration over a time period, use the total change in velocity divided by the total time: a_avg = (v_final − v_initial) / total_time. This gives a single value that represents the overall rate of velocity change, even if acceleration varied at different moments.
Example: if a car goes from 10 m/s to 25 m/s over 6 s with stops and bursts in between, average acceleration = (25 − 10)/6 = 2.5 m/s². This does not describe instant behavior but summarizes the overall change.
High Complexity (Analytical & Scenario-Based)
Q6. A car accelerates from 18 m/s to 27 m/s in 3 s, then brakes from 27 m/s to 9 m/s in 6 s. Calculate the acceleration during each interval, state whether the motions are uniform, and find the overall average acceleration for the 9 s period.
Answer:
First interval: change = 27 − 18 = 9 m/s in 3 s, so a₁ = 9/3 = 3 m/s². This is positive acceleration; if the increase was steady, the motion is uniformly accelerated over that interval.
Second interval: change = 9 − 27 = −18 m/s in 6 s, so a₂ = −18/6 = −3 m/s². This is negative acceleration (deceleration). If braking was steady, that interval is also uniformly accelerated (with negative a).
Overall change in velocity = 9 − 18 = −9 m/s over 9 s, so a_avg = −9/9 = −1 m/s². The overall motion is non-uniform because acceleration changed between intervals, but each interval can be uniform if rates were steady.
Q7. You ride a roller coaster that goes up, then sharply down, then up again. Describe qualitatively how your velocity and acceleration change through these stages and explain why acceleration can be felt even when speed seems constant.
Answer:
As the coaster climbs, speed often decreases so velocity magnitude lowers; acceleration is negative relative to motion direction (slowing). At the top, direction changes rapidly, so velocity vector changes, causing acceleration even if speed is momentarily small.
Going sharply down, speed increases quickly; acceleration is positive in the direction of motion (gravity increases velocity). When moving along curved tracks, acceleration also points toward the curve center (change in direction), so riders feel forces.
Even at constant speed in a curve, velocity direction changes and thus centripetal acceleration exists; we feel this as push into the seat or side. Acceleration is a change in velocity magnitude or direction, so it is experienced even if speed remains constant.
Q8. A train starts from rest and reaches 20 m/s uniformly in 10 s. Find its acceleration and the distance covered in those 10 s. Explain each step and formula used in simple terms.
Answer:
The train starts with u = 0 and reaches v = 20 m/s in t = 10 s. Use a = (v − u)/t: a = (20 − 0)/10 = 2 m/s². This gives a constant acceleration of 2 m/s².
To find distance under uniform acceleration, average velocity = (u + v)/2 = (0 + 20)/2 = 10 m/s. Distance = average velocity × time = 10 × 10 = 100 m.
So the train accelerates at 2 m/s² and covers 100 m during the 10 s. Using average velocity is valid because acceleration is uniform in this problem.
Q9. Explain why acceleration is a vector quantity. Give two distinct examples that show acceleration direction may differ from velocity direction, and describe the physical consequences.
Answer:
Acceleration measures change in velocity; velocity itself is a vector (has magnitude and direction). Therefore acceleration must be a vector because it describes how the velocity vector changes in magnitude or direction.
Example 1: A car braking while moving forward has velocity forward but acceleration backward; consequence: speed decreases and the car slows down.
Example 2: A satellite moving at constant speed in circular orbit has velocity tangent to the circle, while acceleration points toward the center (centripetal). Consequence: direction changes continuously, keeping the satellite in orbit though speed remains nearly constant.
These examples show acceleration can change speed, direction, or both.
Q10. While cycling downhill you increase speed from 6 m/s to 12 m/s in 4 s. Then you apply brakes and reduce speed to 3 m/s in 3 s. Calculate the accelerations for both phases and explain the type of motion in each phase.
Answer:
Downhill speeding: change = 12 − 6 = 6 m/s in 4 s, so a₁ = 6/4 = 1.5 m/s² (positive). This is positive acceleration; if the increase is steady the motion is uniformly accelerated during these 4 s.
Braking phase: change = 3 − 12 = −9 m/s in 3 s, so a₂ = −9/3 = −3 m/s² (negative). This is deceleration; if the slowing is steady, the motion is uniformly accelerated with negative acceleration in this interval.
Overall, two different uniform accelerations occur sequentially, making the full ride non-uniform because acceleration changes between phases.
