Question

The average velocity of a body moving with uniform acceleration after travelling a distance of $3.06m$ is $0.34m/s$. The change in velocity of the body is $0.18m/s$. During this time, its 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)

$V_{avg}=\frac{D}{t}\Rightarrow 0.34=\frac{3.06}{t}$

$\Rightarrow t=\frac{3.06}{0.34}$ = $9sec$

We have, $v=u+at$
$v-u=at$
$0.18=a\times 9$

$\Rightarrow a=\frac{0.18}{9}=0.02m/s^2$