Question

A projectile starts from the origin at time $t=0$ and moves in the $\text{xy-plane}$ with a constant acceleration $\alpha$ in the $\text{y-direction}$ and constant speed along $x$ direction. If equation of motion is $y=\beta x^2$ then its velocity component in the $\text{x-direction}$ will be

• A. $\frac{2\alpha }{\beta }$
• B. $\sqrt{\frac{2\alpha }{\beta }}$
• C. $\frac{\alpha }{2\beta }$
• D. $\sqrt{\frac{\alpha }{2\beta }}$

Answer 1

The correct option is D $\sqrt{\frac{\alpha }{2\beta }}$

Given, equation of motion,
$y=\beta x^2$

Differentiating the equation of motion with respect to time, $t$ we have

$\frac{dy}{dt}=2\beta x\frac{dx}{dt}.....\left(1\right)$

Where,
$\frac{dy}{dt}=v_y$ $($Velcity along $\text{y-direction})$

$\frac{dx}{dt}=v_x$ $($Velcity along $\text{x-direction})$

Substituting the above values in equation $\left(1\right)$, we have

$v_y=2\beta x{v}_x......\left(2\right)$

Differentiating equation $\left(2\right)$,

$\frac{d{v}_y}{dt}=2\beta (v_x\frac{dx}{dt}+x\frac{d{v}_x}{dt})....\left(3\right)$

Since,
$\frac{d{v}_y}{dt}=a_y$ $($ Acceleration along $\text{y-direction})$

$\frac{d{v}_x}{dt}=a_x$ $($ Acceleration along $\text{x-direction})$

From equation $\left(3\right)$,

$\Rightarrow a_y=2\beta \left(\right(v_x{)}^2+x{a}_x)$

According to problem,
Acceleration along $\text{y-direction}$, $a_y=\alpha$ and acceleration along $\text{x-direction}$,$a_x=0$ at time, $t=0$

Putting these in the above equation we get,
$\alpha =2\beta \left(\right(v_x{)}^2+x\times 0)$

$\therefore v_x=\sqrt{\frac{\alpha }{2\beta }}$

Why this question? Concept involved: $\left(1\right)$ Instantaneous velocity $\left(2\right)$ Instantaneous acceleration This question has been asked to brush up your concept of differentiation and it's application in kinematics.