A train is moving with a velocity of $90 k m p h$ . When brakes are applied, the retardation produced is $0.5 m s^{- 2}$ . Find the velocity after $10 s$ .

20 m s^{- 1}

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25 m s^{- 1}

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30 m s^{- 1}

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40 m s^{- 1}

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Solution

The correct option is C $20 m s^{- 1}$
Initial Velocity, $u = 90 k m p h = \frac{90 \times 1000}{60 \times 60} m s^{- 1} = 25 m s^{- 1}$

Retardation, $a = - 0.5 m s^{- 2}$

Time, $t = 10 s$
Using equation of motion
$v = u + a t$

$v = 25 - (0.5 \times 10) = 20 m s^{- 1}$

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A train is moving with a velocity of 90 kmph. When brakes are applied, the retardation produced is 0.5 ms−2. Find the velocity after 10s.

The correct option is C 20ms−1Initial Velocity, u=90 kmph=90×100060×60ms−1=25 ms−1Retardation, a=−0.5 ms−2Time, t=10 sUsing equation of motionv=u+atv=25−(0.5×10)=20 ms−1

A train is moving with a velocity of $90 k m p h$ . When brakes are applied, the retardation produced is $0.5 m s^{- 2}$ . Find the velocity after $10 s$ .

The correct option is C $20 m s^{- 1}$
Initial Velocity, $u = 90 k m p h = \frac{90 \times 1000}{60 \times 60} m s^{- 1} = 25 m s^{- 1}$

Retardation, $a = - 0.5 m s^{- 2}$

Time, $t = 10 s$
Using equation of motion
$v = u + a t$

$v = 25 - (0.5 \times 10) = 20 m s^{- 1}$