Question

A particle after starting from rest, experiences constant acceleration for $20$ sec. If it covers a distance $S_1$ in first $10$ sec, then the distance covered during next $10$ sec will be

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

Answer 1

Answer: (C)

In $20$ seconds, total distance travelled
$s=\frac{1}{2}a(20{)}^2=\frac{1}{2}a\times 400$
$=200a$
In first $10$ seconds, the distance $s_1$ travelled
$s_1=\frac{1}{2}a\times (10{)}^2$
$=50a$
So, in other $10$ seconds, the distance $s_2$ travelled
$s_2=s-s_1$
or $s_2=200a-50a$
or $=150a$
or $=3\times (50a)$
Hence, $s_2=3s_1$