Question

A particle is moving with initial velocity $3\hat{i}+4\hat{j}$ $m/s$ and acceleration of body is $4\hat{i}-3\hat{j}$ $m/s^2$. Then path of the particle is

• A. Circle
• B. Ellipse
• C. Straight line
• D. Parabola

Answer 1

The correct option is D Parabola

Here first we will find dot product.

If a vector $x=A\hat{i}+B\hat{j}andC\hat{i}+D\hat{j}$

Dot product of $x$ and $y=AC+BD$

So $3\times 4-4\times 3=V\times F\times \cos x$

So, $x$ is $90$. This means velocity is perpendicular to acceleration and here acceleration is constant.

This is the case of parabolic path when it is thrown horizontally

Hence it is parabolic.