A particle moving along X -axis has acceleration f , at time t...

& Acceleration

f = \dfrac{dv}{dt} = {f}_{0} \left(1 + \dfrac{t}{T}\right)</p><p>ordv = {f}_{0}\left($1 -$\dfrac{t}{T}\right) \cdot$dt . . . . . . . . . Eq ( i )$

Integrating Eq. $( i )$ on both sides, we get

$v = {f}_{0}t -$\dfrac{f0}{T} \cdot \dfrac{t2}{2} +$C$

After applying boundary conditions~

$v = 0$ at $t = 0$

We get, $C = 0$

$⇒ v = {f}_{0}t -$\dfrac{f0}{T} \cdot \dfrac{t2}{2} . . . . . . . . . .$Eq ( ii )$~ &

As $f = {f}_{0} \left(1 - \dfrac{t}{T}\right)$

When ${f}_{0} = 0, t = T$

Substituting $t = T$~in Eq. $( ii )$, then velocity

${v}_{x} = {f}_{0}T - \dfrac{f0}{T} \cdot \dfrac{T2}{2} = \dfrac{1}{2}{f}_{0}T$