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A car driver driving at a speed of ${90 kmh}^{- 1}$ applies brakes and brings the car to rest in $20 s .$ The acceleration of the car is

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+ 1 m s^{- 2}

- 1.5 m s^{- 2}

- 1.25 m s^{- 2}

+ 2.5 m s^{- 2}

answer is B.

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Detailed Solution

A car driver driving at a speed of

{90 kmh}^{- 1}

applies brakes and brings the car to rest in

20 s .

The acceleration of the car is

- 1.25 m s^{- 1}

.
Given data in the problem,
Initial velocity,

u = 90 km h^{- 1}

Final velocity,

v = 0

Time,

t = 20 s

Converting km

h^{- 1}

to m

s^{- 1}

\Rightarrow 90 km h^{- 1} = 90 \times \frac{5}{18} m s^{- 1}

= 25 m s^{- 1}

By using the first law of motion,

v = u + at

Where,

v

= final velocity

u

= initial velocity

a

= acceleration

t

= time

\Rightarrow 0 = 25 + a \times 20

\Rightarrow a = - 25 / 20

\Rightarrow a = - 1.25 m s^{- 1}

(Negative sign indicates that acceleration is decreasing.)
So, the acceleration of the car after

20 s

of applying brakes is

- 1.25 m s^{- 1}

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