A motorcycle has initial velocity of 5 m/s. After 3 seconds, the velocity is 7 m/s. The displacement of the motorcycle in 3 seconds is:

Correct — B, 18 m. Two routes give the same number, and running both is the safest way to answer a kinematics item under time pressure. The direct route: the velocity rises from 5 m/s to 7 m/s in 3 seconds, so the acceleration is (7 minus 5) divided by 3, that is 2/3 metres per second squared. Substituting into s equals ut plus one-half a t squared gives 5 times 3, which is 15, plus one-half times 2/3 times 9, which is 3 — a total of 18 metres. The shorter route uses the fact that under uniform acceleration the average velocity is simply the mean of the initial and final values: (5 plus 7) divided by 2 is 6 m/s, and 6 metres per second for 3 seconds is 18 metres. Note that the second route needs no acceleration at all, which is why it is worth reaching for first whenever a question supplies both end velocities and the time.

• (a)21 m — This is 7 times 3 — the final velocity treated as if it had applied for the whole interval. The motorcycle was slower than 7 m/s for all but the last instant.
• (c)36 m — This is 12 times 3, using the sum of the two velocities instead of their average. The average of 5 and 7 is 6, not 12.
• (d)6 m — This is the average velocity itself, 6 m/s, read off as a distance. It still has to be multiplied by the 3 seconds.

For motion in a straight line with uniform acceleration, three equations connect the five quantities involved — initial velocity, final velocity, acceleration, time and displacement: v equals u plus at; s equals ut plus one-half a t squared; and v squared equals u squared plus 2as. Each omits one quantity, so the one to choose is whichever leaves out the quantity you neither know nor want. A fourth relation, s equals the average of u and v times t, follows from the first two and is often the quickest of all.

The trap in items of this shape is not the physics but the arithmetic shortcut taken in haste — multiplying the final velocity by the time, or the sum of the velocities by the time. Both mistakes are represented in the options, which is a reliable sign that the examiner is testing carefulness rather than understanding. A useful discipline is to sanity-check the answer against bounds before choosing: the displacement must lie between the distance covered at the slowest speed, 5 times 3 equals 15 metres, and at the fastest, 7 times 3 equals 21 metres. Only 18 metres falls inside that window, which eliminates three options before any equation is written.

• Under uniform acceleration, a = (v - u)/t, so here a = (7 - 5)/3 = 2/3 m/s squared.
• The second equation of motion, s = ut + one-half a t squared, gives 15 + 3 = 18 m.
• For uniform acceleration the average velocity is the arithmetic mean of the initial and final velocities, (u + v)/2, here 6 m/s.
• Displacement equals average velocity multiplied by time, 6 x 3 = 18 m — the same answer by an independent route.
• The displacement must lie between ut = 15 m and vt = 21 m, which rules out every option except 18 m.

• Multiplying the final velocity by the total time, which overstates the displacement.
• Using the sum of the two velocities instead of their average.
• Forgetting that the average-velocity shortcut is valid only when the acceleration is uniform.

As a short numerical supplying three of the five kinematic quantities, or as a graph question asking what the area under a velocity-time line represents.

• practice — not a real PYQ

  A body starting from rest attains a velocity of 12 m/s in 4 seconds under uniform acceleration. Its displacement in those 4 seconds is:

  • (a)48 m
  • (b)24 m
  • (c)12 m
  • (d)6 m

  Answer(b) 24 m — the average velocity is (0 + 12)/2 = 6 m/s, and 6 x 4 = 24 m.

• practice — not a real PYQ

  For a body moving in a straight line with uniform acceleration, the area under its velocity-time graph gives which one of the following?

  • (a)The acceleration
  • (b)The displacement
  • (c)The average speed
  • (d)The momentum

  Answer(b) The displacement — velocity multiplied by time is displacement, and that product is precisely the area under the curve.