Question

An object accelerates from rest to a velocity 27.5 m/s in 10 sec then find distance covered by object in next 10 sec

• A. 550 m
• B. 137.5 m
• C. 412.5 m
• D. 275 m

Answer 1

The correct option is C 412.5 m

u=0,v=27.5m/s and t=10 sec.
$\therefore a=\frac{27.5-0}{10}=2.75m/s^2$
Now,the distance travelled in next 10 sec,
$S=ut+\frac{1}{2}a{t}^2=27.5\times 10+\frac{1}{2}\times 2.75\times 100$
=275+137.5=412.5 m