Question

An object accelerates from rest to a velocity $27.5{\text{ms}}^{-1}$ in $10$ sec. Find the distance covered by the object (in meter)during next $10$ sec.

Answer 1

Given $u=0,v=27.5{\text{ms}}^{-1},t=10s$
$a=\frac{v-u}{t}=\frac{27.5}{10}=2.75{\text{m/s}}^2$
In the first 10 s, distnace travelled:
$s_1=0\times 10+\frac{1}{2}\times 2.75\times {10}^2=137.5\text{m}$
In the first 20 s, distance travelled:
$s_2=0\times 20+\frac{1}{2}\times 2.75\times {20}^2=550\text{m}$
Required distance $=550-137.5=412.5\text{m}$