The speed of a car travelling on a straight road is listed below at successive intervals of 1 s : Time (s) 0 1 2 3 4 Speed (m/s) 0 2 4 6 8 Which of the following is/are correct ? The car travels 1. with a uniform acceleration of 2 m/s². 2. 16 m in 4 s. 3. with an average speed of 4 m/s. Select the correct answer using the code given below :
- (a)1, 2 and 3
- (b)2 and 3 only
- (c)1 and 2 only
- (d)1 only
Correct — A, 1, 2 and 3. Take the three claims in turn against the table. The speed rises from 0 to 2 to 4 to 6 to 8 metres per second, gaining exactly 2 m/s in every one-second interval, so the acceleration is (8 minus 0) divided by 4, which is 2 m/s squared, and it is the same in every interval — that is what uniform acceleration means, so the first claim holds. For the distance, use s = ut + half a t squared with u = 0, a = 2 and t = 4, giving s = half times 2 times 16 = 16 m; the same number comes from the area under the straight-line speed-time graph, a triangle of base 4 s and height 8 m/s, so the second claim holds. Average speed is total distance divided by total time, which is 16 m divided by 4 s = 4 m/s, and for uniform acceleration it also equals half of (initial plus final speed), that is half of (0 plus 8) = 4 m/s. All three statements are therefore correct.
- (b)2 and 3 only — This drops the first statement, but the table shows the speed increasing by the same 2 m/s in each of the four intervals, which is precisely a uniform acceleration of 2 m/s squared. Nothing in the data suggests a changing acceleration.
- (c)1 and 2 only — This drops the third statement, yet the average speed follows directly from the first two — 16 m covered in 4 s is 4 m/s. It is also the mean of the initial and final speeds, which is valid here because the acceleration is uniform.
- (d)1 only — This keeps only the acceleration and rejects both the distance and the average speed, although both were computed from that very acceleration and match the area under the speed-time graph.
When acceleration is uniform, the speed-time graph is a straight line, the three equations of motion apply, and the area under the graph gives the distance covered. Average speed is always total distance divided by total time; only for uniform acceleration does it also equal the arithmetic mean of the initial and the final speed.
A statement set like this is answered fastest by checking each line against a single quantity computed once. Work out a = 2 m/s squared from any interval, then get the distance as the triangular area under the graph, then divide by the time. Students most often lose this question by rejecting the third statement, having half-remembered that average speed cannot be found by averaging speeds — that warning applies to a journey made in two stages at different constant speeds, not to motion with uniform acceleration, where the mean-of-endpoints shortcut is exact.
- Uniform acceleration means equal changes of velocity in equal intervals of time, giving a straight-line speed-time graph.
- Here the acceleration is (8 - 0)/4 = 2 metres per second squared.
- The distance in 4 s is s = ut + half a t squared = 0 + half x 2 x 16 = 16 metres, which is also the area under the graph.
- Average speed is total distance divided by total time = 16/4 = 4 metres per second.
- For uniform acceleration only, average speed also equals half of (initial speed + final speed).
All three statements survive the check, so the answer is 1, 2 and 3.
- Averaging two speeds when the motion is not uniformly accelerated — for a two-stage journey at different constant speeds the average speed is total distance over total time, not the plain mean.
- Reading the area under a distance-time graph as a distance; it is the slope that carries the meaning there.
- Assuming any increase of speed is uniform acceleration without checking that the increase per interval is equal.
NDA sets short numerical items from a small table or a graph — compute the acceleration first, then use the area under the speed-time graph for the distance.
A ball is dropped from the top of a high building with a constant acceleration of 9.8 m/s². What will be its velocity after 3 seconds?
- (a) 9.8 m/s
- (b) 19.6 m/s
- (c) 29.4 m/s
- (d) 39.2 m/s
Answer(c) 29.4 m/s
The same uniform-acceleration relation v = u + at applied to a body starting from rest, which is the first of the three checks this NDA question asks for.
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
States the graphical form of the very motion tabulated here, and the area under that straight line is what gives the 16 m in 4 s.
Ram records the odometer readings of his car for the distance covered from 2000 km at the start of his journey and 2400 km at the end of the journey after 8 hours. What is the average speed of the car ?
- (a) 50 km/h
- (b) 60 km/h
- (c) 70 km/h
- (d) 80 km/h
Answer(a) 50 km/h
Uses the same definition of average speed as total distance over total time that settles the third statement in this question.
- practice — not a real PYQ
A body starts from rest and moves with a uniform acceleration of 4 m/s squared. The distance covered by it in the first 5 seconds is
- (a)20 m
- (b)40 m
- (c)50 m
- (d)100 m
Answer(c) 50 m — with u = 0, s = half a t squared = half x 4 x 25 = 50 metres.
- practice — not a real PYQ
A car uniformly accelerates from 10 m/s to 30 m/s in 10 seconds. Its average speed over this interval is
- (a)10 m/s
- (b)15 m/s
- (c)20 m/s
- (d)30 m/s
Answer(c) 20 m/s — for uniform acceleration the average speed is half of (10 + 30), which is 20 metres per second.