Question 21
A particle moves in a straight line so that its displacement $$x$$ at any time $$t$$ is given by $$x^2 = 1 + t^2$$. Its acceleration at any time $$t$$ is $$x^{-n}$$ where $$n =$$ ___________
Correct Answer: 3
Solution
$$x^2 = 1+t^2$$. $$2x\dot{x} = 2t \Rightarrow \dot{x} = t/x$$.
$$\ddot{x} = (x-t\dot{x})/x^2 = (x-t^2/x)/x^2 = (x^2-t^2)/x^3 = 1/x^3$$.
So $$n = 3$$.
The answer is 3.
Create a FREE account and get:
- Free JEE Mains Previous Papers PDF
- Take JEE Mains paper tests
