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(A) Given acceleration $a = \frac{dv}{dt} = 8 - \frac{t}{5} 	ext{ cm/s}^2$. Integrating with respect to $t$: $v = \int (8 - \frac{t}{5}) dt = 8t - \f

Vedclass · Accepted Answer

$160$

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(A) Given acceleration $a = \frac{dv}{dt} = 8 - \frac{t}{5} 	ext{ cm/s}^2$. Integrating with respect to $t$: $v = \int (8 - \frac{t}{5}) dt = 8t - \frac{t^2}{10} + C$. Since the particle starts from rest,at $t = 0$,$v = 0$. $0 = 8(0) - \frac{0^2}{10} + C \Rightarrow C = 0$. So,the velocity function is $v(t) = 8t - \frac{t^2}{10}$. Acceleration is zero when $8 - \frac{t}{5} = 0$,which gives $t = 40 	ext{ s}$. Substituting $t = 40$ into the velocity equation: $v = 8(40) - \frac{(40)^2}{10} = 320 - \frac{1600}{10} = 320 - 160 = 160 	ext{ cm/s}$.

After $t$ seconds,the acceleration of a particle,which starts from rest and moves in a straight line is $(8-\frac{t}{5}) \text{ cm/s}^2$. The velocity of the particle at the instant when the acceleration is zero is: (in $\text{ cm/s}$)

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