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 $\vec{a}=\frac{d v}{d t}$
Given, $a=\alpha t+\beta t^{2} $
$\Rightarrow \frac{d v}{d t}=\left(\alpha t+\beta t^{2}\right) $
$\Rightarrow d v=\left(\alpha t+\beta t^{2}\right) d t$
Integrating it within the condition of motion
$\int\limits_{v_{0}}^{v} d v=\int\limits_{0}^{t}\left(\alpha t+\beta t^{2}\right) d t $
$\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} $
$\Rightarrow \frac{d x}{d t}=v_{0}+\frac{1}{2} \alpha t^{2}+\frac{1}{3} \beta t^{3}$
Integrating it, using
$\int x^{n} d x=\frac{x^{n+1}}{n+1}$,
we have $\int\limits_{x_{0}}^{x} d x=\int\limits_{0}^{t}\left(v_{0}+\frac{1}{2} \alpha t^{2}+\frac{1}{3} \beta t^{3}\right) d t $
$\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 $x(t)=x_{0}+v_{0} t+\frac{1}{6} \alpha t^{3}+\frac{1}{12} \beta t^{4}$