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 $\left( {{v_a}} \right)$ of the particle is then:
(A) $\dfrac{1}{2}{f_o}T$
(B) ${f_o}T$
(C) $\dfrac{1}{2}{f_o}{T^2}$
(D) ${f_o}{T^2}$

Answer 1

We know that acceleration is a derived physical quantity which can be defined as the rate of change of velocity. Mathematically acceleration is given by, $a = \dfrac{{{v_2} - {v_1}}}{t}$ . It can also be given by $a = \dfrac{{dv}}{{dt}}$ for linear acceleration.
Formula used: We will be using the formulas $a = \dfrac{{{v_2} - {v_1}}}{t} = \dfrac{{dv}}{{dt}}$ where $a$ is the acceleration of the body, ${v_2}$ is the final velocity of the body, ${v_1}$ is the initial velocity of the body, and $t$ is the time taken for the velocity to change.

Complete step by step answer:
We know that acceleration of a body is the rate of change of velocity of the body with respect to time. Mathematically it can also be given as, a derivative of velocity with respect to time.
$a = \dfrac{{dv}}{{dt}}$
From the problem we have acceleration of a particle given by $f$ as a function of time as,
$f = {f_0}\left( {1 - \dfrac{t}{T}} \right)$ . We also know from the problem that the velocity of the particle $v$ at time $t = 0$ is also zero. $t = 0 \Rightarrow v = 0$ .
Let us consider the body has no acceleration or $f = 0$ ,
$0 = {f_0}\left( {1 - \dfrac{t}{T}} \right)$
Simplifying the equation, we get,
$0 = 1 - \dfrac{t}{T}$
$\dfrac{t}{T} = 1 \Rightarrow t = T$
Thus, we now know that at acceleration $f = 0$ , $t = T$ .
We are required to find the velocity of the body at this point, when $f = 0$ . So, we know that $a = \dfrac{{dv}}{{dt}}$ . [Applying this in our function for acceleration]
$\dfrac{{dv}}{{dt}} = {f_0}\left( {1 - \dfrac{t}{T}} \right)$
$dv = {f_0}\left( {1 - \dfrac{t}{T}} \right)dt$
Integrating with respect to $t$ on both sides,
$\int {dv} = \int {{f_0}\left( {1 - \dfrac{t}{T}} \right)dt}$
${v_a} = {f_0}\left( {t - \dfrac{{{t^2}}}{{2T}}} \right)$ [Since $f{}_0$ and $T$ are constants.]
But we know that between the intervals $t = 0$ and $f = 0$, $t = T$. Thus, substituting $t = T$ we get,
${v_a} = {f_0}\left( {T - \dfrac{{{T^2}}}{{2T}}} \right)$
Taking LCM and solving the brackets,
${v_a} = {f_0}\left( {\dfrac{{2{T^2} - {T^2}}}{{2T}}} \right) = {f_0}\left( {\dfrac{{{T^2}}}{{2T}}} \right)$
${v_a} = {f_0}\left( {\dfrac{T}{2}} \right)$
Thus, the velocity of the body at that interval will be ${v_a} = \dfrac{1}{2}{f_0}T$ .

Hence the correct answer is option A.

Note: In kinematics each quantity is related to the other by either being its derivative of integral.
$v = \dfrac{{ds}}{{dt}};a = \dfrac{{dv}}{{dt}}$ , while $v = \int {a.dt;s = \int {v.dt} }$ .
This relation can be better understood when a graph is plotted with either displacement, velocity or acceleration on the y-axis and time on the x-axis.