Consider the following velocity and time graph : Which one of the following is the value of average acceleration from 8 s to 12 s?
- (a)8 m/s²
- (b)12 m/s²
- (c)2 m/s²
- (d)–1 m/s²
Answer
Why
Correct — D, –1 m/s². The printed graph has velocity in m/s up the vertical axis and time in seconds along the horizontal, marked at 4, 8, 12 and 16 s. The trace holds flat at 8 m/s from the start until t = 8 s, then falls in a straight line to 4 m/s at t = 12 s, and stays flat at 4 m/s afterwards.
Average acceleration over an interval is the change in velocity divided by the time taken, a = (v_final − v_initial) / (t_final − t_initial). Between 8 s and 12 s the velocity goes from 8 m/s down to 4 m/s, a change of −4 m/s, over 4 s.
a = (4 − 8) / (12 − 8) = −4/4 = −1 m/s². The sign is part of the answer: the body is slowing, so the acceleration is negative — it opposes the motion.
Why the others are wrong
- (a)8 m/s² — This is the starting velocity read straight off the graph, not an acceleration. A velocity in m/s cannot be an acceleration in m/s²; the interval's change in velocity still has to be divided by the 4 s it took.
- (b)12 m/s² — This is the time coordinate 12 s carried over as if it were the answer. Neither endpoint of the interval is the acceleration.
- (c)2 m/s² — Right in size only if the velocity had dropped by 8 m/s over 4 s, or if the 4 m/s change were divided by 2 s. The change here is 4 m/s over 4 s, giving 1 m/s² in magnitude — and it must carry a minus sign because the body is slowing.
Concept
On a velocity–time graph the gradient is the acceleration and the area under the trace is the distance covered. A flat section means constant velocity and therefore zero acceleration; a straight sloping section means uniform acceleration; a downward slope means the body is slowing, which is a negative acceleration, or retardation.
The item is really testing whether you read the interval the question names rather than the graph as a whole. From 0 to 8 s the acceleration is zero and after 12 s it is zero again — only the 8 s to 12 s segment slopes. The other frequent slip is dropping the minus sign: a magnitude of 1 m/s² is not offered on its own, so the sign is what separates the right option from a near miss.
Key facts
- Average acceleration = change in velocity / time taken.
- On a v–t graph, acceleration is the gradient of the line.
- A flat v–t line means zero acceleration, not zero velocity.
- A falling line means negative acceleration (retardation).
- Here: (4 − 8) m/s over (12 − 8) s = −1 m/s².
Only the 8–12 s segment slopes; the change of −4 m/s over 4 s gives −1 m/s², option (d).
Study next
Common traps
- Reading a velocity off the axis and offering it as an acceleration.
- Ignoring the negative sign when the body is slowing.
- Using the whole graph instead of the interval the question names.
NDA prints a multi-segment v–t graph and names one interval — take the two end velocities of that interval only, divide by its duration, and keep the sign.
Related PYQs
What is the nature of velocity-time graph for a car moving with uniform acceleration?
- (a) Parabola
- (b) Logarithmic
- (c) Straight line
- (d) Exponential
Answer(c) Straight line
The same graph read the other way round — it asks what shape a v–t graph takes under uniform acceleration, which is the straight sloping segment used here.
Practice
- practice — not a real PYQ
On a velocity–time graph, a straight line sloping downwards to the right indicates
- (a)uniform velocity
- (b)uniform acceleration in the direction of motion
- (c)uniform retardation
- (d)the body is at rest
Answer(c) uniform retardation — a constant negative gradient means the velocity falls steadily. - practice — not a real PYQ
A body's velocity falls uniformly from 20 m/s to 5 m/s in 5 s. What is its average acceleration?
- (a)+3 m/s²
- (b)−3 m/s²
- (c)−15 m/s²
- (d)−5 m/s²
Answer(b) −3 m/s² — (5 − 20)/5 = −3 m/s², negative because the body is slowing.