Question

A particle is moving in a straight line with initial velocity and uniform acceleration a. If the sum of the distance travelled in $t^{\text{th }}$ and $(t+1{)}^{\text{th }}$ seconds is $100,cm$, then its velocity after t seconds, in $cm/s$, is

• A. 80
• B. 50
• C. 20
• D. 30

Answer 1

Answer: (B)

The distance travelled in $n^{\text{th }}$ second is
$S_n=u+\frac{1}{2}(2n-1)a...(1)$
So distance travelled in $t^{\text{th}}\&(t+1{)}^{\text{th }}$ second are
$S_t=u+\frac{1}{2}(2t-1)a...(2)$
$S_{t+1}=u+\frac{1}{2}(2t+1)a...(3)$
As per question,
$S_t+S_{t+1}=100=2(u+at)...(4)$
Now from first equation of motion the velocity of particle after time $t$,
if it moves with an acceleration $a$ is
$v=u+at...(5)$
where $u$ is initial velocity
So from equation (4) and (5),
we get $v=50cm/s$