Question

An object accelerates from rest to a velocity $27.5m/s$ in $10\sec$ then the distance covered in next $10\sec$ is

• A. 550 m
• B. 137.5 m
• C. 412.5 m
• D. 175 m

Answer 1

Answer: (C)

Given : $u=0,v=27.5m/s,t=10\sec$
From first equation of metion
$v=u+at$
$27.5=0+a\times 10$
$a=2.75m/s^2$
Distance covered in first $10\sec$ is
$s_1=ut+\frac{1}{2}a{t}^2$
$=0\times 10+\frac{1}{2}\times 2.75\times (10{)}^2$
$=\frac{1}{2}\times 2.75\times 100$
$=137.5m$
Distance covered in next $10\sec$
with uniform velocity of $27.5m/s$
$s_2=27.5\times 10=275m$
Total distance covered is
$=137.5+275=412.5m$