A motor car moving with a uniform speed of $20 m s^{- 1}$ comes to rest on the application of brakes after travelling a distance of 10 m. Its acceleration is

$20 m / s e c^{2}$

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$- 20 m / s e c^{2}$

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$- 40 m / s e c^{2}$

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$+ 2 m / s e c^{2}$

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Solution

The correct option is B
$- 20 m / s e c^{2}$

Using, $v^{2} = u^{2} + 2 a S$

$\Rightarrow o = u^{2} + 2 a S$

$\Rightarrow a = - \frac{u^{2}}{2 S}$

$a = - \frac{(20)^{2}}{2 \times 10}$

$a = - 20 m s^{- 2}$

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A motor car moving with a uniform speed of 20 ms−1 comes to rest on the application of brakes after travelling a distance of 10 m. Its acceleration is

The correct option is B −20m/sec2 Using, v2=u2+2aS ⇒o=u2+2aS ⇒a=−u22S a=−(20)22×10 a=−20ms−2

A motor car moving with a uniform speed of $20 m s^{- 1}$ comes to rest on the application of brakes after travelling a distance of 10 m. Its acceleration is

The correct option is B
$- 20 m / s e c^{2}$

Using, $v^{2} = u^{2} + 2 a S$

$\Rightarrow o = u^{2} + 2 a S$

$\Rightarrow a = - \frac{u^{2}}{2 S}$

$a = - \frac{(20)^{2}}{2 \times 10}$

$a = - 20 m s^{- 2}$