Question

A particle moving along the x-axis has acceleration $f$ at time $t$, given by $f=f_0\left(1-{\displaystyle \frac{t}{T}}\right)$, 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 velocity ($v_x$) of the particle is then:

Answer 1

Step 1: Given Data

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

$t=0$, $v=0\Rightarrow f=0$ then we have $0=f_0\left(1-{\displaystyle \frac{t}{T}}\right)$

Step 2: Formula used
Acceleration is defined the rate of change in velocity :

$a={\displaystyle \frac{dv}{dt}}\Rightarrow dv=adt\Rightarrow \int dv=\int adt$
Step 3: Calculations
The equation of motion of this particle is given as:
Acceleration $f=f_0\left(1-{\displaystyle \frac{t}{T}}\right)$
The time interval between $t=0$ and the instant when $f=0$.

We require to find the instant at which$f=0$.
At $t=0$, $v=0\Rightarrow f=0$ then we have $0=f_0\left(1-{\displaystyle \frac{t}{T}}\right)$
Since $f_0$ is a constant T$\Rightarrow f_0\ne 0\Rightarrow \left(1-{\displaystyle \frac{t}{T}}\right)=0\Rightarrow 1={\displaystyle \frac{t}{T}}\Rightarrow t=T$.
At the instant $t=T$ the acceleration$f=0$.
Acceleration,

$f={\displaystyle \frac{dv}{dt}}\Rightarrow dv=fdt$
Thus integrating the above equation within the limits of $t=0$ and $t=T$, $v_x$ is the velocity of the particle,
${\int }_{v=0}^{v=v_x}dv={\int }_{t=0}^{t=T}fdt={\int }_0^T{f}_0\left(1-{\displaystyle \frac{t}{T}}\right)dt={\displaystyle \frac{f_0}{T}}{\int }_0^T(T-t)dt$
$\Rightarrow v_x={\displaystyle \frac{f_0}{T}}{\left[Tt-{\displaystyle \frac{t^2}{2}}\right]}_0^T={\displaystyle \frac{f_0}{T}}\left[T^2-{\displaystyle \frac{T^2}{2}}\right]={\displaystyle \frac{f_0}{T}}{\displaystyle \frac{T^2}{2}}={\displaystyle \frac{1}{2}}{f}_{0}T$

Hence, In the time interval between $t=0$ and the instant when $f=0$the average velocity ($v_x$) of the particle is $\frac{1}{2}{f}_{0}T$.