EPFO APFC 2023 General Studies · Question 53 of 120

Correct — B, (b) 4500 m. Three quantities are given and one is asked for, and the whole question is decided by converting the units before doing anything else. The train starts from rest, so the initial velocity u is zero. The final speed is 108 km/h, and a speed in kilometres per hour becomes one in metres per second on multiplying by 5/18, so v = 108 × 5/18 = 30 m/s. The time is 5 minutes, which is 300 seconds. The quickest route needs no acceleration at all. Under uniform acceleration the average velocity over any interval is the mean of the initial and final velocities, so it is (0 + 30)/2 = 15 m/s, and the distance is the average velocity multiplied by the time: 15 × 300 = 4500 m. The standard route gives the same figure. The acceleration is a = (v − u)/t = 30/300 = 0·1 m/s², and then s = ut + ½at² = 0 + ½ × 0·1 × 300² = ½ × 0·1 × 90000 = 4500 m. The third equation of motion is another way home: s = (v² − u²)/2a = 900/0·2 = 4500 m. All three agree, which is the check worth making when a question can be attacked from more than one direction. The single most valuable habit here is the unit conversion. Leave the speed in kilometres per hour and the time in minutes and the answer comes out in units that match no option; convert both at the start and everything after it is arithmetic. The factor 5/18 is worth memorising in both directions — multiply by 5/18 to go from km/h to m/s, and by 18/5 to come back — and it is worth noticing that 108 km/h is 30 m/s, a pair of numbers that recurs constantly in examination questions.

• (a)9000 m — Exactly twice the correct answer, and therefore the option that catches the commonest error in the whole of kinematics: treating the motion as though it happened at the final speed throughout. Multiply 30 m/s by 300 s and 9000 m is what comes out. But the train begins at rest and only reaches 30 m/s at the end of the five minutes, so it spends the whole interval travelling more slowly than that — its average speed is 15 m/s, half the final speed, which is precisely why the answer is half of 9000. The same error appears as dropping the factor of one half from ½at², and it is worth building the habit of asking whether a computed distance is consistent with the average speed rather than the peak.
• (c)355 m — This value does not follow from any correct application of the equations of motion, and it can be dismissed by a size check that needs no algebra. Even at the average speed of 15 m/s, a distance of 355 m would be covered in about 24 seconds, whereas the train is accelerating for a full 300 seconds — more than twelve times as long. Any answer that small is impossible for a vehicle that reaches 108 km/h. The general defence against options of this kind is to estimate the answer before computing it: with an average speed of the order of 15 m/s over 300 s, the distance must be of the order of thousands of metres, which leaves only two of the four printed values in play.
• (d)108 m — This simply reprints the number given in the stem with its unit changed from kilometres per hour to metres, which is the oldest distractor in the examiner's collection and works on candidates who are copying figures rather than computing with them. The number 108 is a speed in this question, not a distance, and the two are not interchangeable whatever the arithmetic. The size check disposes of it even faster than of the previous option: a train accelerating for five minutes to a speed of 30 m/s cannot possibly have covered only 108 m, a distance it would cross in less than four seconds at its final speed.

Motion in a straight line with constant acceleration is described completely by three equations, each linking a different set of the five quantities involved — initial velocity, final velocity, acceleration, time and displacement. The first, v = u + at, leaves out the displacement; the second, s = ut + ½at², leaves out the final velocity; the third, v² = u² + 2as, leaves out the time. Choosing the equation that omits the quantity you neither know nor want is what makes these problems quick. A fourth relation is often the fastest of all and is sometimes forgotten: under uniform acceleration the average velocity equals the mean of the initial and final velocities, so the displacement is simply that mean multiplied by the time. It follows from the shape of the velocity-time graph, where the displacement is the area under the line and the area of a trapezium is the average of the parallel sides times the base — and reading these problems off a velocity-time graph, rather than from a remembered formula, is a habit worth acquiring, because the graph makes the factor of one half visible rather than something to be recalled. Everything else in this family is unit discipline. Speeds are given in kilometres per hour and times in minutes or hours, while the answers are wanted in metres and seconds, and the conversion factor between the two speed units is 5/18 in one direction and 18/5 in the other.

The general science block of this paper carries one short computation among its conceptual items, and this is it — a single application of uniformly accelerated motion, with the difficulty deliberately placed in the unit conversion rather than in the physics. The design of the option set says as much: the value twice the correct one is offered for the candidate who forgets that the train started from rest, and two far smaller values are offered for the candidate who mishandles the units or copies a number out of the stem. Both of the small options are eliminated by an order-of-magnitude estimate before any equation is written, which makes this a good item on which to practise estimating first and computing second. For a paper with negative marking at one-third of the mark for each wrong answer, the ability to reduce four options to two in ten seconds is worth as much as the ability to compute, because it turns a blind guess into a favourable one. Note the printed form: the stem is an incomplete sentence completed by each option, with no question mark; fifty of this paper's stems end in neither a question mark nor a colon, and this is one of them.

• To convert kilometres per hour into metres per second multiply by 5/18; 108 km/h is exactly 30 m/s, and 5 minutes is 300 s.
• Under uniform acceleration the average velocity is the mean of the initial and final velocities, so distance = mean velocity × time = 15 × 300 = 4500 m.
• The acceleration here is a = (v − u)/t = 30/300 = 0·1 m/s², and s = ut + ½at² = ½ × 0·1 × 300² = 4500 m.
• The third equation of motion gives the same result without the time: s = (v² − u²)/2a = 900/0·2 = 4500 m.
• On a velocity-time graph the displacement is the area under the line, which for motion from rest is a triangle — the geometric reason for the factor of one half.
• Distance computed as final speed multiplied by time is correct only for uniform velocity; for motion starting from rest it overstates the answer by a factor of two.

• Using the final speed as though it were the constant speed of the whole journey, which doubles the answer for motion starting from rest
• Leaving the speed in kilometres per hour or the time in minutes; both must be converted before any equation is applied
• Dropping the factor of one half in s = ut + ½at²
• Choosing an option because the number appears in the stem — 108 is a speed here, not a distance
• Failing to estimate first; an average speed of 15 m/s over 300 s puts the answer in the thousands of metres and eliminates half the option set instantly

Kinematics appears in this paper as a one-step numerical item embedded in the general science block, with the units deliberately mismatched between the stem and the options. Expect the distractor at twice or half the correct value, expect one option copied from the stem, and expect the arithmetic itself to be easy once the conversion is done. The reliable method is to convert every quantity to metres and seconds first, then choose the equation that omits what you do not need.

No directly related past PYQ was found.

• practice — not a real PYQ

  A car moving at 72 km/h is brought to rest by uniform braking in 10 seconds. The distance covered while stopping is

  • (a)200 m
  • (b)100 m
  • (c)72 m
  • (d)40 m

  Answer(b) 100 m — convert first: 72 km/h × 5/18 = 20 m/s. The motion ends at rest, so the average velocity is (20 + 0)/2 = 10 m/s, and the distance is 10 × 10 = 100 m. Equivalently the retardation is 2 m/s² and s = (v² − u²)/2a gives 400/4 = 100 m. Using the initial speed throughout would give 200 m, twice the correct value.

• practice — not a real PYQ

  A body starting from rest moves with a uniform acceleration of 2 m/s². The ratio of the distance it covers in the first 4 seconds to the distance it covers in the first 2 seconds is

  • (a)2 : 1
  • (b)3 : 1
  • (c)4 : 1
  • (d)8 : 1

  Answer(c) 4 : 1 — for a body starting from rest the distance is ½at², which is proportional to the square of the time, so doubling the time quadruples the distance. Numerically the distances are 16 m and 4 m. The same proportionality explains why the distances covered in successive equal intervals from rest are in the ratio 1 : 3 : 5 : 7.