A particle moving along $x$-axis has acceleration $f$, at time $t$, given by $f=f_{0}\left(1-\frac{t

Question

A particle moving along $x$-axis has acceleration $f$, at time $t$, given by $f=f_{0}\left(1-\frac{t^2}{T^2}\right)$, 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 $\left(v_{x}\right)$ is

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(b) : Given : At time $t=0$, velocity, $v=0$. Acceleration $f=f_{0}\left(1-\frac{t^2}{T^2}ight)$ \ At $f=0, \quad 0=f_{0}\left(1-\frac{t^2}{T^2}ight) \implies t=T$. \ Also, acceleration $f=\frac{d v}{d t}$ \ $	herefore \quad \int_{0}^{v_{x}} d v=\int_{0}^{T} f_{0}\left(1-\frac{t^2}{T^2}ight) d t$ \ $	herefore \quad v_{x}=\left[f_{0} t-\frac{f_{0} t^{3}}{3 T^2}ight]_{0}^{T}=f_{0} T-\frac{f_{0} T^{3}}{3 T^2}=f_{0} T - \frac{f_{0} T}{3} = \frac{2}{3} f_{0} T$.