Question

A point moves in xy plane, according to law x=kt and y= kt(1- $\alpha$t) where k and $\alpha$ are positive constants and t is time. Find the time after which the angle between the velocity and acceleration is ${45}^{\circ }$

• A. $\frac{1}{\alpha }$
• B. $\frac{2}{\alpha }$
• C. $\frac{3}{\alpha }$
• D. $\frac{4}{\alpha }$

Answer 1

The correct option is A $\frac{1}{\alpha }$

$x=kt,y=kt-k\alpha t^2$

$v_x=\frac{dx}{dt}=k$ ,$v_y=\frac{dy}{dt}=k-2k\alpha t$

$a_x=\frac{d^{2}x}{d{t}^2}=0,a_y=\frac{d^{2}y}{d{t}^2}=-2k\alpha$

Since $a_x$ is zero and $a_y$ is $-2k\alpha$

$a$ is $-2k\alpha$ is y-axis direction.

For the angle between v and a to be ${45}^{\circ }$ we must have:$v_x=-v_y$

$k=-k+2k\alpha t$

$2k\alpha t=2k$

$t=\frac{1}{\alpha }$