If the velocity of a particle is (10 + 2t^2) | vedclass

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(D) Average acceleration is defined as the change in velocity divided by the time interval.$a_{avg} = \frac{v(t_2) - v(t_1)}{t_2 - t_1}$Given $v(t) =

Vedclass · Accepted Answer

$14$

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(D) Average acceleration is defined as the change in velocity divided by the time interval.$a_{avg} = \frac{v(t_2) - v(t_1)}{t_2 - t_1}$Given $v(t) = 10 + 2t^2$,$t_1 = 2 \ s$,and $t_2 = 5 \ s$.Calculate velocity at $t_1 = 2 \ s$: $v(2) = 10 + 2(2)^2 = 10 + 8 = 18 \ m/s$.Calculate velocity at $t_2 = 5 \ s$: $v(5) = 10 + 2(5)^2 = 10 + 50 = 60 \ m/s$.Now,calculate average acceleration:$a_{avg} = \frac{60 - 18}{5 - 2} = \frac{42}{3} = 14 \ m/s^2$.

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

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