Question

The average velocity of a body moving with uniform acceleration travelling a distance of 3.06 m is 0.34 ms^–1. If the change in velocity of the body is 0.18ms^–1 during this time, its uniform acceleration is

• A. 0.01 ms^–2
• B. 0.02 ms^–2
• C. 0.03 ms^–2
• D. 0.04 ms^–2

Answer 1

Answer: (B)

0.02 ms^–2

Answer 2

$\text{Time} = \frac{\text{Distance}}{\text{Average velocity}} = \frac{3.06}{0.34} = 9sec$

and

$\text{Acceleration} = \frac{\text{Change in velocity}}{\text{Time}} = \frac{0.18}{9} = 0.02m/s^{2}$.