Question

A particle travels $10m$ in first $5s$, $10m$ in next $3s$. Assuming constant acceleration, what is the distance traveled in the next two seconds.

• A. $\frac{20}{3}m$
• B. $20m$
• C. $60m$
• D. $180m$

Answer 1

The correct option is A $\frac{20}{3}m$

In the first $5$ sec:

$s=ut+0.5a{t}^2$

$10=5u+0.5a\times 25$

In the next $3$ sec:

$20=8u+0.5a\times 64$

Solving above two equations will give us:

$u=\frac{7}{6},a=\frac{1}{3}$

Therefore, in next $2$ sec distance traveled is:

$S=\frac{7}{6}\times 10+0.5\times \frac{1}{3}\times 100=\frac{170}{6}$

Therefore distance travelled in next $2$ sec is $=\frac{170}{6}-20=\frac{20}{3}$