Question

The average velocity of a body moving with uniform acceleration after travelling a distance of 3.06m is $ 0.34m{{s}^{-1}}. $ If the change in velocity of the body is $ 0.18\text{ }m{{s}^{-}}1, $ during this time its uniform acceleration is:

• A. $ 0.01m{{s}^{-2}} $
• B. $ 0.02m{{s}^{-2}} $
• C. $ 0.03m{{s}^{-2}} $
• D. $ 0.04m{{s}^{-2}} $

Answer 1

Answer: (B)

Time taken in covering a distance of 3.06 m is $ \text{t =}\frac{\text{Distance}}{\text{Average velocity}}=\frac{3.06}{0.34}=9s $ Also, change in velocity $ v-u=0.18\,m{{s}^{-1}} $ From first law of motion $ v=u+at $ $ \Rightarrow $ $ a=\frac{v-u}{t}=\frac{0.18}{9}=0.02\,m{{s}^{-2}} $