Question

A particle starting from rest moves upto $20\, s$ with constant acceleration. If $S _1$ is the distance covered in first $10\, s$ and $S_2$ is the distance covered in last $10$ second, then

• A. $S_2=S_1$
• B. $S_2=2S_1$
• C. $S_2=3S_1$
• D. $S_2=4S_1$

Answer 1

Answer: (C)

$S_2=3S_1$

Answer 2

$S_1 =\frac{1}{2} at^2 = \frac{1}{2} \times a \times 10^32 = 50a$ $S_1 + S_2 = \frac{1}{2} \times a \times 20^2 = 200 a$ Then $S_2 = 20 \, a - S_1 = 200 \, a - 50 \, a = 150 \, a = 3 \times 50 \, a = 3S_1$