If displacement of a particle is given as $x = t^{3} - 2 t^{2}$ , then average acceleration from $0$ to $2$ seconds is

2 m / s^{2}

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4 m / s^{2}

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8 m / s^{2}

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1 m / s^{2}

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Solution

The correct option is A $2 m / s^{2}$
$v = \frac{d x}{d t} = 3 t^{2} - 4 t$
$v_{t = 2} = 4 m / s$ & $v_{t = 0} = 0 m / s$
$< a >= \frac{Δ V}{Δ t} = \frac{4 - 0}{2} = 2 m / s^{2}$

Suggest Corrections

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