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Q.

If a cyclist pedalling at a speed of ${5 ms}^{- 1}$ picks up a speed of ${10 ms}^{- 1}$ over a distance of 50 m, then his acceleration is

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a

${9 ms}^{- 2}$

b

${0.75 ms}^{- 2}$

c

${7.5 ms}^{- 2}$

d

${3 ms}^{- 2}$

answer is A.

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Detailed Solution

$v^{2} - u^{2} = 2 as$

$\Rightarrow {(10)}^{2} - {(5)}^{2} = 2 \times a \times 50$

$\Rightarrow 100 - 25 = 10 0a$

$\Rightarrow a = \frac{75}{100} = 0.7 {5m/s}^{2}$

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