Physics - Kinematics Question with Solution | TestHub

 Acceleration

$f=\frac{dv}{dt}=f_0\left(1+\frac{t}{T}\right)$

or $dv=f_0\left(1-\frac{t}{T}\right)·dt ......... Eq\left(i\right)$

Integrating Eq. $\left(\mathrm{i}\right)$ on both sides, we get

$v=f_{0}t-\frac{f_0}{T}·\frac{t^2}{2}+C$

After applying boundary conditions 

$v=0$ at $t=0$

We get, $C=0$

$\Rightarrow v=f_{0}t-\frac{f_0}{T}·\frac{t^2}{2} .......... Eq\left(ii\right)$   

As $f=f_0\left(1-\frac{t}{T}\right)$

When $f_0=0,t=T$

Substituting $t=T$ in Eq. $\left(\mathrm{ii}\right)$, then velocity

$v_x=f_{0}T-\frac{f_0}{T}·\frac{T^2}{2}=\frac{1}{2}{f}_{0}T$