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

Answer: (C)

412.5 m

Answer 2

u = 0, $v = 27.5\mspace{6mu} m/s$ and t = 10 sec

$\therefore a = \frac{27.5 - 0}{10} = 2.75\mspace{6mu} m/s^{2}$

Now, the distance traveled in next 10 sec,

$S = ut + \frac{1}{2}at^{2} = 27.5 \times 10 + \frac{1}{2} \times 2.75 \times 100$

= 275 + 137.5 = 412.5 m