The driver of a train traveling at $40 m s^{- 1}$ applies the brakes as the train enters a station. The train slows down at a rate of $2 m s^{- 2}$ . The distance traveled by train before stopping is:

200 m

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300 m

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400 m

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500 m

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Solution

The correct option is C $400 m$
Assumption: Initial direction of velocity is positive.
Given:
Initial speed, $u = 40 m s^{- 1}$
Acceleration, $a = - 2 m s^{- 2}$
Acceleration is negative because it is opposite to the direction of initial velocity.
As the train finally stops, final velocity, $v = 0$

Let the distance traveled by train before stopping be $s$ .

Using equation of motion,
$v^{2} = u^{2} + 2 a s$
Substituting values,
$0^{2} = 40^{2} + (2 \times - 2 \times s)$
Solving, we get:
$s = + 400 m$

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The driver of a train traveling at 40 m s−1 applies the brakes as the train enters a station. The train slows down at a rate of 2 m s−2. The distance traveled by train before stopping is:

The driver of a train traveling at $40 m s^{- 1}$ applies the brakes as the train enters a station. The train slows down at a rate of $2 m s^{- 2}$ . The distance traveled by train before stopping is: