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The breaks applied to a car produce an acceleration of $6 m s^{- 2}$ 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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15 m

6 m

12 m

17 m

answer is C.

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Detailed Solution

The distance it travels during this time is

12 m

.
Given data in the problem,
Acceleration of car,

a = - 6 m s^{- 2}

(We are using negative sign because the car is moving opposite to the motion.)
Final velocity,

v = 0

Time,

t = 2 s

By using the first equation of motion,

v = u + at

Where,

v = final velocity

u = initial velocity

a = acceleration

t = time

0 = u + (- 6) \times 2

{u = 12 ms}^{- 1}

Now, to calculate the distance travelled by car,
By using the second equation of motion,

s = ut + \frac{1}{2} {at}^{2}

Where,

s = distance

u = initial velocity

a = uniform acceleration

t = time

putting all values in the above equation,

= > s = 12 \times 2 + \frac{1}{2} \times {(- 6) \times 2}^{2}

= > s = 24 - \frac{1}{2} \times 6 \times 4

= > s = 24 - 12

= > s = 12 m

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