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