If the velocity of a particle depends upon time as $v (t) = (10 + 2 t^{3}) m / s$ , then the average acceleration of the particle between the times 2s and 5s is

$87 m / s^{2}$

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$260 m / s^{2}$

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$26 m / s^{2}$

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$78 m / s^{2}$

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Solution

The correct option is D
$78 m / s^{2}$

Average acceleration = $\frac{c h a n g e i n v e l o c i t y}{t i m e t a k e n}$

= $\frac{v_{2} - v_{1}}{t_{2} - t_{1}}$

= $\frac{[10 + 2 (5)^{3}] - [10 + 2 (2)^{3}]}{3} = \frac{260 - 26}{3} = \frac{234}{3} = 78 m / s^{2}$

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If the velocity of a particle depends upon time as v(t)=(10+2t3)m/s, then the average acceleration of the particle between the times 2s and 5s is

The correct option is D 78m/s2 Average acceleration = change in velocitytime taken = v2−v1t2−t1 = [10+2(5)3]−[10+2(2)3]3=260−263=2343=78m/s2

If the velocity of a particle depends upon time as $v (t) = (10 + 2 t^{3}) m / s$ , then the average acceleration of the particle between the times 2s and 5s is

The correct option is D
$78 m / s^{2}$

Average acceleration = $\frac{c h a n g e i n v e l o c i t y}{t i m e t a k e n}$

= $\frac{v_{2} - v_{1}}{t_{2} - t_{1}}$

= $\frac{[10 + 2 (5)^{3}] - [10 + 2 (2)^{3}]}{3} = \frac{260 - 26}{3} = \frac{234}{3} = 78 m / s^{2}$