A car is moving in a straight line with speed $18 k m h^{- 1}$ . It is stopped in $5 s$ by applying the brakes. The speed of car after 2 s of applying the brakes is:

10.8 m s^{- 1}

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- 10.8 m s^{- 1}

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3 m s^{- 1}

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- 7 m s^{- 1}

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Solution

The correct option is C $3 m s^{- 1}$
Speed $= 18 k m / h = 5 m / s$
$R e t a r d a t i o n = \frac{- 5 m / s}{5 s} = - 1 m / s^{2}$
$R e t a r d a t i o n = \frac{v_{f} - v_{i}}{t}$
$- 1 m / s^{2} = \frac{v_{f} - 5}{2}$
Therefore, $v_{f} = 3 m / s$

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A car is moving in a straight line with speed 18 km $h^{- 1}$ . It is stopped in 5 s by applying the brakes. Find:

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(ii) the retardation of car
(iii) the speed of car after 2 s of applying the brakes.

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A car of mass 480 kg moving at a speed of 54 km $h^{- 1}$ is stopped by applying brakes in 10 s. Calculte the force applied by the brakes.

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