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Q.

The brakes applied to a car produced an acceleration of $6$ ${ms}^{- 2}$ in the opposite direction to the motion. If the car takes $2 s$ to stop after the application of brakes, calculate the distance it travels during this time.

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a

$12 m$

b

$13 m$

c

$14 m$

d

$15 m$

answer is A.

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Detailed Solution

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Given:
$a = - 6 {ms}^{- 2}, t = 2 s and v = 0 {ms}^{- 1}$
By using the velocity-time relation,
$v = u + at$
$0 = u + (- 6 {ms}^{- 2}) \times 2 s$
$u = 12 {ms}^{- 1}$
By using Position-time relation distance travelled,

$s = ut + \frac{1}{2} {at}^{2}$
$= (12 {ms}^{- 1}) \times (2 s) + \frac{1}{2} (- 6) {(2)}^{2}$

$= 24 m - 12 m$
$= 12 m$

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