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Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of $6 m s^{- 1}$ in $30 s$ . Then, he applies brakes such that the velocity of the bicycle comes down to ${4 m s}^{- 1}$ in the next $5 s$ . Calculate the acceleration of the bicycle when Rahul applies brakes.

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0.4 m s^{- 2}

- 0.8 m s^{- 2}

- 0.4 m s^{- 2}

0.5 m s^{- 2}

answer is C.

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Detailed Solution

The acceleration of the bicycle when Rahul applies brakes is

- 0.4 m s^{- 2}

.
Given data in problem,
When Rahul applies brakes
Initial velocity,

u = 6 m s^{- 1}

Final velocity,

v = 4 m s^{- 1}

Time,

t = 5 s

So,
By using the acceleration formula,

Acceleration = \frac{change in velocity}{time}

\Rightarrow a = \frac{v - u}{t}

\Rightarrow a = \frac{4 - 6}{5}

\Rightarrow a = \frac{- 2}{5}

\Rightarrow a = - 0.4 m s^{- 2}

So, the acceleration of bicycle when Rahul applies break is

- 0.4 m s^{- 2}

. Negative sign shows the retardation and cycle is slowing down.

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