Q 11-03-013NEETAIPMT 2011Top questionEasy
A body is moving with velocity $30\ \text{m s}^{-1}$ towards east. After $10$ s its velocity becomes $40\ \text{m s}^{-1}$ towards north. The average acceleration of the body is
Answer: (C) $5\ \text{m s}^{-2}$
Average acceleration uses the change in the velocity vector:
$$\Delta\vec{v} = 40\,\hat{N} - 30\,\hat{E}, \qquad |\Delta\vec{v}| = \sqrt{40^2 + 30^2} = 50\ \text{m s}^{-1}$$
$$a_{avg} = \frac{50}{10} = 5\ \text{m s}^{-2}$$
(Subtracting the speeds, $40 - 30 = 10$, would be wrong: the direction changed.)
Solution by Sreeraj P, M.Sc Physics