Question

A particle moving along $x-$axis has acceleration $f$, at time $t$, given by $f=f_0(1-\frac{t}{T}),$ 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:

• A. $\frac{1}{2}{f}_{0}T$
• B. $f_{0}T$
• C. $\frac{1}{2}{f}_0{T}^2$
• D. $f_0{T}^2$

Answer 1

Answer: (A)

$\frac{1}{2}{f}_{0}T$

let any time t particle has a velocity u and in a change in time (t+dt) its final velocity is u+dv, this dt time acceleration is constant

using first eq

$v=u+at$

$u+du=u+f_0(1-\frac{t}{T})dt$

${\int }_0^{v}du={\int }_0^T{f}_0(1-\frac{t}{T})dt$

$v=f_0{(t-\frac{t^2}{2T})}_0^T$

$v=f_0(T-\frac{T}{2})$

$v=f_0\frac{T}{2}$

Hence the option A correct