Question

Instantaneous velocity of an object (starting from origin) varies with time as $v=\alpha -\beta t^2$(where $\alpha$ and $\beta$ are positive constants). If $x$ and $a$ are the position and acceleration of the object as a function of time. Then,

• A. $x=\alpha t-\frac{1}{3}\beta t^3$
• B. $x=\alpha t-\frac{1}{2}\beta t^2$
• C. $\alpha =-\beta t$
• D. Maximum positive displacement from origin $=\frac{2\alpha }{3}\sqrt{\frac{\alpha }{\beta }}$

Answer 1

Answer: (A), (D)

The correct options are
A $x=\alpha t-\frac{1}{3}\beta t^3$
D Maximum positive displacement from origin $=\frac{2\alpha }{3}\sqrt{\frac{\alpha }{\beta }}$

Acceleration of the object is given as
$a=\frac{dv}{dt}=\frac{d}{dt}(\alpha -\beta t^2)=-2\beta t$
For displacement we use
$v=\frac{dx}{dt}=\alpha -\beta t^2......\text{(i)}$
Integrating with proper limits, we get
$\underset{0}{\overset{x}{\int }}dx=\underset{0}{\overset{t}{\int }}(\alpha -\beta t^2)$
Or $x=\alpha t-\frac{1}{3}\beta t^3......\text{(ii)}$
At $t=0,v=\alpha$
Velocity is positive in the beginning and acceleration is negative, so the velocity will keep on decreasing until it becomes $0$ and then it reverses its direction.
Displacement in the positive direction will be maximum when $v$ becomes zero.
$0=\alpha -\beta t^2$
$\Rightarrow t=\sqrt{\frac{\alpha }{\beta }}......\text{(iii)}$
Putting this in equ. $\text{(ii)}$, we get
$x_{max}=\alpha \sqrt{\frac{\alpha }{\beta }}-\frac{1}{3}\beta {\left(\frac{\alpha }{\beta }\right)}^{3/2}$
$\Rightarrow$Maximum positive displacement from the origin
$\Rightarrow x_{max}=\frac{2}{3}\alpha \sqrt{\frac{\alpha }{\beta }}$