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NVAEasyJEE 2023Velocity & Acceleration

JEE Physics 2023 Question with Solution

For a train engine moving with a speed of 20 m s−120 \, \text{m s}^{-1}, the driver must apply brakes at a distance of 500 m500 \, \text{m} before the station for the train to come to rest at the station. If the brakes were applied at half of this distance, the train engine would cross the station with speed x m s−1\sqrt{x} \, \text{m s}^{-1}. The value of xx is _____. (Assuming the same retardation is produced by brakes)

Answer

Correct answer:200

Step-by-step solution

Standard Method

Given: Initial speed u=20 m/su = 20 \, \text{m/s}, stopping distance S1=500 mS_1 = 500 \, \text{m}, final speed at station v=0v = 0.

Find: The value of xx if the brakes are applied at half the distance and the train crosses the station with speed x m/s\sqrt{x} \, \text{m/s}.

Using the third equation of motion,

v2=u2+2aSv^2 = u^2 + 2aS

For the first case,

0=(20)2+2a(500)0 = (20)^2 + 2a(500) 0=400+1000a0 = 400 + 1000a a=−4001000=−0.4 m/s2a = -\frac{400}{1000} = -0.4 \, \text{m/s}^2

The negative sign shows retardation.

Now the brakes are applied at half the distance, so

S2=5002=250 mS_2 = \frac{500}{2} = 250 \, \text{m}

Again using

v2=u2+2aS2v^2 = u^2 + 2aS_2 v2=(20)2+2(−0.4)(250)v^2 = (20)^2 + 2(-0.4)(250) v2=400−200v^2 = 400 - 200 v2=200v^2 = 200 v=200 m/sv = \sqrt{200} \, \text{m/s}

Since the speed is given as x m/s\sqrt{x} \, \text{m/s}, we get

x=200x = 200

Therefore, the value of xx is 200200.

Common mistakes

  • Using different retardation values in the two cases. The question clearly states that the same retardation is produced by brakes, so the same aa must be used in both applications of the equation of motion.

  • Ignoring the sign of acceleration. Since the train is slowing down, acceleration is negative. Taking aa as positive gives an incorrect value of v2v^2 in the second part.

  • Using the full 500 m500 \, \text{m} again in the second case instead of half the distance. The brakes are applied at half of the original distance, so the correct distance is 250 m250 \, \text{m}.

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