A particle moving along x-axis has accelerati | vedclass

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Question

$\begin{array}{l}
Give\,:\,At\,time\,t = 0,\,velocity,\,v = 0\
Accelerations\,f\, = \,{f_0}\left( {1 - \frac{t}{T}} ight)\
At\,f = 0,\,\,\,\,0

Vedclass · Accepted Answer

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

Vedclass · Answer

$\begin{array}{l}
Give\,:\,At\,time\,t = 0,\,velocity,\,v = 0\
Accelerations\,f\, = \,{f_0}\left( {1 - \frac{t}{T}} ight)\
At\,f = 0,\,\,\,\,0 = {f_0}\left( {1 - \frac{t}{T}} ight)\
\sin ce\,{f_0}\,is\,a\,constant,\
	herefore \,1 - \frac{t}{T} = 0\,\,\,or\,\,t = T.
\end{array}$
$\begin{array}{l}
Also,\,acceleation\,f = \frac{{dv}}{{dt}}\
	herefore \,\int\limits_0^{{v_x}} {dv}  = \int\limits_{t = 0}^{t = T} {fdt = \int\limits_0^T {{f_0}} } \left( {1 - \frac{t}{T}} ight)dt\
	herefore \,\,{v_x} = \left[ {{f_0}t - \frac{{{f_0}{t^2}}}{{2T}}} ight]_0^T = {f_0}T - \frac{{{f_0}{T^2}}}{{2T}} = \frac{1}{2}{f_0}T
\end{array}$

A particle moving along $x-$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 $(v_x)$ is

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