Question

Velocity and acceleration of a particle are
$V=(2\hat{i}-4\hat{j})m/s$ and $a=(-2\hat{i}+4\hat{j})m/s^2$ Which type if motion is this?

Answer 1

Given, $u=(2\hat{i}-4\hat{j})m{s}^{-1}anda=(-2\hat{i}+4\hat{j})m{s}^{-2}$

Apply kinematic equation of motion.

$v=u+at$

$v=(2\hat{i}-4\hat{j})+(-2\hat{i}+4\hat{j})t$

$v=(2-2t)\hat{i}-(4-4t)\hat{j}$

It is clear from above expression that velocity will decrease with time.

Hence, It is a retardation motion.