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

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Question

(d) Average acceleration $ = \frac{{{\rm{Change\, in\, velocity }}}}{{{\rm{Time\, taken }}}}$

$ = \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}$ $ = \fra

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(d) Average acceleration $ = \frac{{{\rm{Change\, in\, velocity }}}}{{{\rm{Time\, taken }}}}$

$ = \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}$ $ = \frac{{\left[ {10 + 2{{(5)}^2}} \right] - \left[ {10 + 2{{(2)}^2}} \right]}}{3} = \frac{{60 - 18}}{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 $2s$ and $5s$ is..........$m/s^2$

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