Question

Instantaneous velocity of an object varies with time as $v=\alpha -\beta t^2$. Find its position, $x$ as a function of time, $t$. Also find the object`s maximum positive displacement, $x_{max}$ from the origin.

• A. $x=\alpha t-\frac{\beta t^3}{3}$ and $x_{max}=\frac{2}{3}\frac{{\alpha }^{3/2}}{{\beta }^{3/2}}$
• B. $x=\alpha t-\frac{\beta t^3}{3}$ and $x_{max}=\frac{2\alpha }{{\beta }^{1/2}}$
• C. $x=2\alpha t-\beta$ and $x_{max}=2\alpha \beta$
• D. $x=2{\alpha }^{2}t-\beta t^3$ and $x_{max}=\frac{2{\alpha }^{1/2}}{{\beta }^{3/2}}$

Answer 1

The correct option is A $x=\alpha t-\frac{\beta t^3}{3}$ and $x_{max}=\frac{2}{3}\frac{{\alpha }^{3/2}}{{\beta }^{3/2}}$

Given,
velocity of the object, $v=\alpha -\beta t^2$

From the velocity and displacement relation, we get

$v=\frac{dx}{dt}=\alpha -\beta t^2$

$\Rightarrow dx=(\alpha -\beta t^2)dt$

Integrating within proper limits, we have

${\int }_o^{x}dx={\int }_o^t(\alpha -\beta t^2)dt$

$\Rightarrow x=\alpha t-\frac{1}{3}\beta t^3...\left(1\right)$

To calculate the maximum value of $x$ i.e., $x_{max}$,by differentiating the above equation w.r.t $t$ and equating it to zero, we get

$\frac{dx}{dt}=\alpha -\beta t^2=0$

Now, to check whether $x$ is maximum or not at $t=\sqrt{\frac{\alpha }{\beta }}$, we have to differente $\frac{dx}{dt}$ again and substitute the value of $t$.

$\frac{d^{2}x}{d{t}^2}=-2t\beta$

$\therefore {\left[\frac{d^{2}x}{d{t}^2}\right]}_{t=\sqrt{\frac{\alpha }{\beta }}}=-2\sqrt{\alpha \beta }$

Since, the value of $\frac{d^{2}x}{d{t}^2}$ is negative at $t=\sqrt{\frac{\alpha }{\beta }}$, so the $x$ will have maximum value at this instant.

Thus, the maximum value of $x$,

$x_{max}=\alpha \sqrt{\frac{\alpha }{\beta }}-\frac{\beta }{3}{\left(\sqrt{\frac{\alpha }{\beta }}\right)}^3$

$\Rightarrow x_{max}=\frac{{\alpha }^{3/2}}{{\beta }^{1/2}}-\frac{{\alpha }^{3/2}}{3.{\beta }^{1/2}}$

$\therefore x_{max}=\frac{2}{3}\frac{{\alpha }^{3/2}}{{\beta }^{1/2}}$

Hence, option (a) is the correct answer.