Question

A particle is moving along x-axis has acceleration $f$ at time $t$, given by $f =f_{0}\left(1 - \dfrac{t}{T}\right)$, 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:
A. $\dfrac{1}{2}f_0T$
B. $f_0T$
C. $\dfrac{1}{2}f_0T^2$
D. $f_0T^2$

Answer 1

First deduce the time or the instant at which $f =0$ using the equation of motion given to us in the question. Remember that $f_0$ is a constant and will not possess a null value. Then once you are able to obtain that particular instant, you can now integrally find the value of this velocity in the required time period. In other words, defining acceleration as the rate of change of velocity, arrive at an expression for the particle velocity in terms of acceleration that you can integrate over time.

Formula used:
Acceleration as the rate of change of velocity : $a =\dfrac{dv}{dt} \Rightarrow dv = a\;dt \Rightarrow \int dv = \int a\;dt$

Complete answer:
The equation of motion of this particle is given as:
Acceleration $f =f_{0}\left(1 - \dfrac{t}{T}\right)$
We are first required to define the time interval between $t=0$ and the instant when $f=0$. Specifically, we need to find the instant at which $f=0$.
Given that at $t=0$, $v =0 \Rightarrow f =0$, then we have $0 =f_{0}\left(1 - \dfrac{t}{T}\right)$
Now, since $f_0$ is a constant $\Rightarrow f_0 \neq 0 \Rightarrow \left(1 - \dfrac{t}{T}\right) = 0 \Rightarrow 1 = \dfrac{t}{T} \Rightarrow t =T$.
This means that at the instant $t =T$ the acceleration $f =0$.
Alternatively, the acceleration can be expressed as the rate of change of velocity, i.e., $f =\dfrac{dv}{dt} \Rightarrow dv = f\;dt$
Thus we can integrate the above equation within the limits of $t=0$ and $t=T$ to obtain the required velocity of the particle:
$\int_{v=0}^{v=v_x} dv = \int_{t=0}^{t=T} f\;dt = \int_{0}^{T} f_{0}\left(1 - \dfrac{t}{T}\right) dt = \dfrac{f_0}{T} \int_{0}^{T}(T-t)dt$
$\Rightarrow v_x = \dfrac{f_0}{T}\left[ Tt - \dfrac{t^2}{2}\right]_{0}^{T} = \dfrac{f_0}{T}\left[ T^2 - \dfrac{T^2}{2}\right] = \dfrac{f_0}{T}\dfrac{T^2}{2} = \dfrac{1}{2}f_0T$

So, the correct answer is “Option A”.

Note:
It is important to remember that velocity is the integral of acceleration over time. This can easily be deduced from the fact that acceleration is the rate of change of velocity. Recall that the slope of a velocity-time graph will give you the acceleration whereas the area under the curve of an acceleration-time graph gives the velocity.
Also, while integrating, ensure that you are well aware of which terms are constants and which terms will get acted upon by the integral. Therefore, do not forget to identify the appropriate integrands and establish proper limits in case of definite integrals.