If the velocity of a particle is $(10 + 2 t^{2})$ , then the average acceleration of the particle in $m / s^{2}$ between 2s and 5s is :

2

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4

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12

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14

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Solution

The correct option is D $14$
Given,

$v = (10 + 2 t^{2}) m / s$

$t_{1} = 2 s e c$

$t_{2} = 5 s e c$

$v_{1} = 10 + 2 \times (2)^{2}$ (at $t_{1} = 2 s e c$ )

$v_{1} = 18 m / s$

$v_{2} = 10 + 2 \times (5)^{2}$ ( at $t = 5 s e c$ )

$v_{2} = 60 m / s$

The average acceleration, $a = \frac{v_{2} - v_{1}}{t_{2} - t_{1}}$

$a = \frac{60 - 18}{5 - 2}$

$a = 14 m / s^{2}$

The correct option is D.

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If the velocity of a particle is (10+2t2), then the average acceleration of the particle in m/s2 between 2s and 5s is :

The correct option is D 14Given,v=(10+2t2)m/st1=2sect2=5secv1=10+2×(2)2 (at t1=2sec)v1=18m/sv2=10+2×(5)2 ( at t=5sec)v2=60m/sThe average acceleration, a=v2−v1t2−t1a=60−185−2a=14m/s2The correct option is D.

If the velocity of a particle is $(10 + 2 t^{2})$ , then the average acceleration of the particle in $m / s^{2}$ between 2s and 5s is :