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

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(C) Given: At time $t = 0$,velocity $v = 0$.Acceleration $f = f_0(1 - t/T)$.At the instant when $f = 0$,we have $0 = f_0(1 - t/T)$. Since $f_0$ is a c

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

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(C) Given: At time $t = 0$,velocity $v = 0$.Acceleration $f = f_0(1 - t/T)$.At the instant when $f = 0$,we have $0 = f_0(1 - t/T)$. Since $f_0$ is a constant,$1 - t/T = 0$,which implies $t = T$.We know that acceleration $f = \frac{dv}{dt}$,so $dv = f dt$.Integrating both sides from $t = 0$ to $t = T$:$\int_{0}^{v_x} dv = \int_{0}^{T} f_0(1 - t/T) dt$$v_x = f_0 \left[ t - \frac{t^2}{2T} \right]_{0}^{T}$$v_x = f_0 \left( T - \frac{T^2}{2T} \right) = f_0 \left( T - \frac{T}{2} \right) = \frac{1}{2} f_0 T$.

$A$ particle moving along the $x-$axis has acceleration $f$ at time $t$,given by $f = f_0(1 - t/T)$,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 $(v_x)$ is:

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