Question

A particle is subjected to acceleration $a=\alpha t+\beta t^2$ , where $\alpha$ and $\beta$ are constants. The position and velocity of the particle at $t=0$ are $x_0$ and $v_0$ respectively. The expression for position of particle at time $c$ is :

• A. $x(t)=x_0+v_{0}t+\frac{1}{6}\alpha t^3+\frac{1}{12}\beta t^4$
• B. $x(t)=x_0+v_{0}t+\frac{1}{6}\alpha t^2+\frac{1}{24}\beta t^3$
• C. $x(t)=x_0+v_{0}t+\frac{1}{12}\alpha t^2+\frac{1}{6}\beta t^3$
• D. $x(t)=x_0+v_{0}t+\frac{1}{6}\alpha t^2+\frac{1}{12}\beta t^3$

Answer 1

Answer: (A)

Rate of change of velocity gives acceleration $(\vec{a})$ of the particle.
That is $\vec{a}=\frac{dv}{dt}$
Given, $a=\alpha t+\beta t^2$
$\Rightarrow \frac{dv}{dt}=\left(\alpha t+\beta t^2\right)$
$\Rightarrow dv=\left(\alpha t+\beta t^2\right)dt$
Integrating it within the condition of motion
$\underset{v_0}{\overset{v}{\int }}dv=\underset{0}{\overset{t}{\int }}\left(\alpha t+\beta t^2\right)dt$
$\Rightarrow v-v_0=\frac{1}{2}\alpha t^2+\frac{\beta t^2}{3}$
$\Rightarrow v=v_0+\frac{1}{2}\alpha t^2+\frac{1}{3}\beta t^3$
$\Rightarrow \frac{dx}{dt}=v_0+\frac{1}{2}\alpha t^2+\frac{1}{3}\beta t^3$
Integrating it, using
$\int x^{n}dx=\frac{x^{n+1}}{n+1}$,
we have $\underset{x_0}{\overset{x}{\int }}dx=\underset{0}{\overset{t}{\int }}\left(v_0+\frac{1}{2}\alpha t^2+\frac{1}{3}\beta t^3\right)dt$
$\Rightarrow x-x_0=v_{0}t+\frac{1}{6}\alpha t^3+\frac{1}{12}\beta t^4$
$\Rightarrow x=x_0+v_{0}t+\frac{1}{6}\alpha t^3+\frac{1}{12}\beta t^4$
Or $x(t)=x_0+v_{0}t+\frac{1}{6}\alpha t^3+\frac{1}{12}\beta t^4$