JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Kinematics

For a train engine moving with speed of

20 ms^{- 1}

, the driver must apply brakes at a distance of 500 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 ms^{- 1}

. The value of x is ______ (Assuming same retardation is produced by brakes)

Answer: 200

Solution

With the same retardation throughout, apply

v^{2} = u^{2} - 2 a s

. Bringing the train to rest over

500

m gives

0 = 2 0^{2} - 2 a (500)

, so

a = 0.4 m s^{- 2}

. Applying the brakes over half the distance,

250

m, the speed at the station satisfies

v^{2} = 2 0^{2} - 2 (0.4) (250) = 400 - 200 = 200

. Since this speed is

x

, the value of

x

200

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Integer answer

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