A particle moving along the X-axis has accele | vedclass

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(C) The acceleration of the particle is given by $f = f_0 \left(1 - \frac{t}{T}\right)$.Since $f = \frac{dv}{dt}$, the velocity $v$ is obtained by int

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$\frac{1}{2} f_0 T$

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(C) The acceleration of the particle is given by $f = f_0 \left(1 - \frac{t}{T}\right)$.Since $f = \frac{dv}{dt}$, the velocity $v$ is obtained by integrating the acceleration with respect to time:$v = \int f dt = \int f_0 \left(1 - \frac{t}{T}\right) dt = f_0 \left(t - \frac{t^2}{2T}\right) + C$.Given that at $t = 0$, $v = 0$, we find the constant of integration $C = 0$.Thus, the velocity at any time $t$ is $v = f_0 \left(t - \frac{t^2}{2T}\right)$.The acceleration becomes zero when $f = 0$, which implies $1 - \frac{t}{T} = 0$, so $t = T$.Substituting $t = T$ into the velocity equation:$v = f_0 \left(T - \frac{T^2}{2T}\right) = f_0 \left(T - \frac{T}{2}\right) = \frac{f_0 T}{2}$.

$A$ particle moving along the $X$-axis has acceleration $f$ at time $t$ given by $f=f_0\left(1-\frac{t}{T}\right)$, where $f_0$ and $T$ are constants. The particle at $t=0$ has zero velocity. In the time interval between $t=0$ and the instant when $f=0$, the particle's velocity is

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