Question

The acceleration of a particle starting from rest varies with time according to relation \[a = \alpha t + \beta \] . The velocity of the particle after a time $t$ will be
A. $\dfrac{{\alpha {t^2}}}{2} + \beta $
B. $\dfrac{{\alpha {t^2}}}{2} + \beta t$
C. $\alpha {t^2} + \dfrac{1}{2}\beta t$
D. $\dfrac{{(\alpha {t^2}) + \beta }}{2}$

Answer 1

Hint: Here we have given a particle which was initially at rest and then varies with the time and we have to calculate the velocity of particle after a time $t$ ,as we know that acceleration is the rate of change of velocity with time and the given expression is in terms of time so by integrating it, we can find v at particular time t.

Formula used:
$a = \dfrac{{dv}}{{dt}}$
Where, $a$ is the acceleration, $\dfrac{{dv}}{{dt}}$ is a very small change in velocity and time.

Complete step by step answer:
We know that acceleration is given as rate of change of velocity with respect to time which is given as,
$a = \dfrac{{dv}}{{dt}}$
Where $\dfrac{{dv}}{{dt}}$ is the very small change in velocity and time.
Now according to the question \[a = \alpha t + \beta \].So,
$\dfrac{{dv}}{{dt}} = \alpha t + \beta \\
\Rightarrow dv = (\alpha t + \beta )dt $

Now on integrating both the sides, since a particle starts from rest, its initial velocity is zero.Therefore, we will integrate from $0$ to $v$
$\int\limits_0^v {dv} = \int\limits_0^t {(\alpha t + \beta )dt} \\
\Rightarrow v = \dfrac{{\alpha {t^2}}}{2} + \beta t \\ $
Hence the correct option is B.

Note: If we didn’t consider the two separate functions for a while, just consider the positive function \[a = \alpha t + \beta \] as at time $t$ seconds it is represented by a positive function then the answer will be wrong. Because here we are integrating from the time when a particle was at rest to the time when it undergoes acceleration. Acceleration of a particle is the time derivative of its velocity. Hence, the velocity of the particle at any instant would be the integral of its acceleration.