Question

An object accelerates from rest to a velocity $27.5\,m/s$ 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)

412.5 m

Answer 2

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