CDS (I) 2023 General Knowledge · Question 48 of 120

Correct — A, 12 m. Take the initial speed as u = 12 m/s, the acceleration as a = −6 m/s² because the car is slowing, and the time as t = 2 s. Check first that the numbers are consistent: v = u + at gives 12 − 6 × 2 = 0, so the car does indeed come to rest exactly at the end of the two seconds, as the stem says. Now use s = ut + ½at², which gives 12 × 2 + ½ × (−6) × 4 = 24 − 12 = 12 m. Two independent routes confirm it. The relation v² = u² + 2as with v = 0 gives 0 = 144 − 12s, so s = 12 m. And since the deceleration is uniform, the average speed over the stop is simply the mean of the initial and final speeds, (12 + 0)/2 = 6 m/s, which over 2 s covers 12 m. The trap is planted in the second option: 24 m is what you get by multiplying the initial speed by the time and forgetting that the car is slowing down the whole way. A braking car covers exactly half the distance a car holding its speed would.

• (b)24 m — The distance the car would cover if it kept moving at 12 m/s for the full two seconds. It ignores the braking altogether, and it is exactly twice the right answer because uniform deceleration to rest halves the distance.
• (c)36 m — Neither the braked nor the unbraked distance. It comes out of mixing the numbers — for instance adding a spurious term rather than subtracting the ½at² correction.
• (d)48 m — Four times the correct value. It would follow from doubling the time or from adding 24 and 24, and it is far beyond what a car losing all its speed in two seconds can travel.

For motion in a straight line with uniform acceleration the three equations of motion apply: v = u + at, s = ut + ½at², and v² = u² + 2as. Deceleration is simply negative acceleration and is handled by giving a a negative sign rather than by changing the equations. On a velocity-time graph the motion here is a straight line falling from 12 m/s to zero over 2 s, and the distance travelled is the area under that line — a triangle of base 2 and height 12, which is 12 m.

The safest habit in a braking numerical is to check the given data for consistency before using it. Here u/|a| = 12/6 = 2 s, which matches the stated stopping time, so the numbers hang together and any answer above 24 m can be discarded at once. The graphical route is the fastest of all: draw the velocity-time line and take the area, which for a triangle is half base times height. It is also worth carrying the physical consequence, because the same relation governs road safety. The braking distance goes as the square of the speed through v² = u² + 2as, so doubling a vehicle's speed quadruples the distance it needs to stop.

• The three equations of uniformly accelerated motion are v = u + at, s = ut + ½at² and v² = u² + 2as.
• Deceleration is negative acceleration; here a = −6 m/s² with u = 12 m/s.
• The stopping time follows from v = u + at with v = 0, giving t = u divided by the magnitude of the deceleration, that is 2 s.
• Under uniform acceleration the average velocity is the mean of the initial and final velocities, here 6 m/s.
• The area under a velocity-time graph gives the displacement, and for this motion the graph is a triangle of area 12 m.

• Using the initial speed for the whole interval, which produces the doubled answer.
• Entering the deceleration as positive and getting a distance larger than the unbraked one.
• Skipping the consistency check between the given deceleration and the given stopping time.

As a braking-distance numerical, as a find-the-stopping-time item, or as a graphical item on the area under a velocity-time curve.

• practice — not a real PYQ

  A body moving at 20 m/s is brought to rest by a uniform deceleration of 5 m/s². The distance it covers before stopping is

  • (a)20 m
  • (b)40 m
  • (c)80 m
  • (d)100 m

  Answer(b) 40 m — from v² = u² + 2as with v = 0, 0 = 400 − 10s, so s = 40 m; the stop takes 4 s.

• practice — not a real PYQ

  If the speed of a vehicle is doubled while the braking deceleration stays the same, its stopping distance becomes

  • (a)the same
  • (b)twice as long
  • (c)three times as long
  • (d)four times as long

  Answer(d) four times as long — stopping distance follows u² divided by twice the deceleration, so it grows with the square of the speed.