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(C) The force on the object is $F = k - \beta t^2$.The acceleration of the particle is given by $a = \frac{F}{m} = \frac{k - \beta t^2}{m}$.The accele

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$17$

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(C) The force on the object is $F = k - \beta t^2$.The acceleration of the particle is given by $a = \frac{F}{m} = \frac{k - \beta t^2}{m}$.The acceleration is zero when $k - \beta t^2 = 0$,which implies $t^2 = \frac{k}{\beta} = \frac{8}{2} = 4$,so $t = 2 \, s$.Since $a = \frac{dv}{dt}$,we have $dv = \frac{k - \beta t^2}{m} \, dt$.Integrating from $t = -2 \, s$ to $t = 2 \, s$:$\int_{v_i}^{v_f} dv = \int_{-2}^{2} \frac{k - \beta t^2}{m} \, dt$$v_f - (-15) = \frac{1}{2/3} \int_{-2}^{2} (8 - 2t^2) \, dt = \frac{3}{2} \left[ 8t - \frac{2t^3}{3} \right]_{-2}^{2}$$v_f + 15 = \frac{3}{2} \left[ (16 - 16/3) - (-16 + 16/3) \right] = \frac{3}{2} \left[ 32 - 32/3 \right] = \frac{3}{2} \left( \frac{64}{3} \right) = 32$$v_f = 32 - 15 = 17 \, m/s$.

$A$ particle of mass $2/3 \, kg$ with velocity $v = -15 \, m/s$ at $t = -2 \, s$ is acted upon by a force $F = k - \beta t^2$. Here,$k = 8 \, N$ and $\beta = 2 \, N/s^2$. The motion is one-dimensional. Then,the speed at which the particle acceleration is zero again,is ........... $m/s$.

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