Question

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

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

Answer 1

Answer: (B)

Average velocity $=\frac{D i s p l a c e m e n t}{t i m e}$
$0.34=\frac{3.06}{t i m e}$
$time =\frac{3.06}{0.34}sec$
$Acceleration =\frac{c h ange ⁡ in ⁡ v e l o c i t y}{t i m e}$
$Acceleration =\frac{0.18}{\left(\frac{3.06}{0.34}\right)}$
$=0.02 \, m \, s^{- 2}$