A body is moving with a velocity of $30 ext{ m/s}$ towards the east. After $10 ext{ s}$, its veloc

Question

A body is moving with a velocity of $30	ext{ m/s}$ towards the east. After $10	ext{ s}$, its velocity becomes $40	ext{ m/s}$ towards the north. What is the magnitude of the average acceleration of the body?

Professor PC Thomas Classes · Answer

$\vec{v}_1 = 30\hat{i}$, $\vec{v}_2 = 40\hat{j}$. The change in velocity is $\Delta\vec{v} = 40\hat{j} - 30\hat{i}$. The magnitude of change in velocity is $|\Delta\vec{v}| = \sqrt{(-30)^2 + (40)^2} = 50	ext{ m/s}$. Average acceleration $= \frac{|\Delta\vec{v}|}{\Delta t} = \frac{50}{10} = 5	ext{ m/s}^2$.