Question

A particle 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 particle’s velocity ${v_x}$is:
(A) $\dfrac{1}{2}{f_0}T$
(B) ${f_0}T$
(C) $\dfrac{1}{2}{f_0}{T^2}$
(D) ${f_0}{T^2}$

Answer 1

Hint The velocity is defined as the rate of change of the position of an object with respect to the frame of reference, and it is a function of time. Velocity is equivalent to the specification of an object’s speed and direction of the motion. Based on this concept we have to solve this question.

Complete step by step answer:
Given $f = {f_0} = \left( {1 - \dfrac{t}{T}} \right)\;$
Where ${f_0}$and T = constant
At $t = 0$and $v = 0$
When $f = 0$, than $0={{f}_{0}}\left( 1-\dfrac{t}{T} \right)$
$\;T = t - - - - (1)$
Acceleration $f = \dfrac{{final - initial\;velocity}}{{Time}}$
$\dfrac{{dv}}{{dt}} = {f_0}\left( {1 - \dfrac{t}{T}} \right)$

$f\;dv = \int {{f_0}\left( {1 - \dfrac{t}{T}} \right)dt}$
$v = {f_0}\left( {t - \dfrac{{{t^2}}}{{2T}}} \right)$
From the equation 1 we get t = T
$v = \dfrac{1}{2}{f_0}T$

Hence, the correct answer is Option A.

Note Velocity gives us an idea about the direction of the movement of the body or the object. It is the prime indicator of the position as well as the rapidness of the object. It is different from speed, which is a scalar quantity.