Q 11-03-161JEE MainJEE Main 2019 (9 Apr, Shift 2)Easy
The position vector of a particle changes with time according to the relation $\vec r(t) = 15t^2\hat i + (4 - 20t^2)\hat j$. What is the magnitude of the acceleration at $t = 1$?
Answer: (D) $50$
$$\vec a = \frac{d^2\vec r}{dt^2} = 30\hat i - 40\hat j \Rightarrow |\vec a| = \sqrt{30^2 + 40^2} = 50$$
(The acceleration is constant, so this holds at $t = 1$ too.)
Solution by Sreeraj P, M.Sc Physics