Courses for Kids
Free study material
Offline Centres
Store Icon

The initial velocity of a particle is $u$(at ${\text{t = 0}}$ ) and the acceleration $a$ is given by $\alpha {t^{\dfrac{3}{2}}}$ . Which of the following relations is valid?
A) $V = u + \alpha {t^{\dfrac{3}{2}}}$
B) $V = u + \dfrac{{3r{t^3}}}{2}$
C) $V = u + \dfrac{2}{5}\alpha {t^{\dfrac{5}{2}}}$
D) $V = u + \alpha {t^{\dfrac{5}{2}}}$

Last updated date: 20th Jun 2024
Total views: 54.6k
Views today: 2.54k
54.6k+ views
Hint: In this question, we are asked to find out the valid relation. We use the simple formula of the acceleration and integrate that equation with applying proper limits. As acceleration is a function of time, we get the correct relation.

Formula used:
$\vec a = \dfrac{{dv}}{{dt}}$
Here, \[{{\vec a}}\;\]is the acceleration,
${\text{dv}}$ is the velocity (small)
${\text{dt}}$ is the time (small)

Complete step by step answer:
Let’s define all the terms which are given in the question.
We have given the initial velocity of the particle i.e. $u$ at$\left( {{\text{t = 0}}} \right)$.
The acceleration is given by $\alpha {t^{\dfrac{3}{2}}}$.
That is, $\vec a = \alpha {t^{\dfrac{3}{2}}}$
Here, the acceleration is dependent on time
Hence, we can say acceleration is a function of time.
Now, from the formula of acceleration,
That is,
$ \Rightarrow $ $\vec a = \dfrac{{dv}}{{dt}}$
$ \Rightarrow \vec adt\; = \;dv\;$ ………………………………… (1)
We know, acceleration is a function of time. The velocity is changing from ${\text{u}}$ to ${\text{v}}$ at time interval ${\text{t = 0}}$ to ${\text{t = t}}{\text{.}}$
Hence, integrating equation (1) with applying these values as limit, we get
$\int_0^t {adt} = \int_u^v {dv} $
Now, we are substituting the value of ${\text{a}}$,
We get,
$ \Rightarrow \int_0^t {\alpha {t^{\dfrac{3}{2}}}dt = \int_u^v {dv} } $
$ \Rightarrow \alpha \left[ {\dfrac{{{t^{\dfrac{3}{2} + 1}}}}{{\dfrac{3}{2} + 1}}} \right]_0^t = \left[ V \right]_u^v$
$ \Rightarrow \alpha \dfrac{2}{5}{t^{\dfrac{5}{2}}} = v - u$
So, $v - u\; = \;\dfrac{2}{5}\alpha {t^{\dfrac{5}{2}}}$
$v = \;u + \dfrac{2}{5}\alpha {t^{\dfrac{5}{2}}}$
This is the final relation for velocity.

Hence, the final answer is option (C) i.e. $v = \;u + \dfrac{2}{5}\alpha {t^{\dfrac{5}{2}}}$.

Additional Information:
$a = \dfrac{{\Delta v}}{{\Delta t}}$ Where $\Delta v = {v_f} - {v_i}$ and $\Delta t\; = {t_f} - {t_i}$.
The equation for acceleration may be written as
$a = \dfrac{{\Delta v}}{{\Delta t}} = \dfrac{{{v_f} - {v_i}}}{{{t_f} - {t_i}}}\;\_\_\left( 1 \right)$
Note that, ${t_f} - {t_i}$ is always positive as ${t_f} > \;{t_i}$ and it is called ‘elapsed time’.
Equation $\left( 1 \right)$ becomes
$a = \dfrac{{{v_f} - {v_i}}}{t}$

Note: Velocity is defined as the rate of change of distance with respect to time, whereas acceleration is defined as the rate of change of Speed. And both of them are vector quantities. To solve these types of questions, simply use the equation of motion, & if the quantity is changing at $t = 0$ to $t = t,$ then integrate it with respect to proper limits. This way you are able to solve such Questions.