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Question

(b) 
Sum of distance travelled in $t^{t h}$ and $(t+1)^{t h}$ seconds is $100 cm$
$u+\frac{a}{2}(2 t-1)+u+\frac{a}{2}(2(t+1)-1)=100$
$2 u+2 a

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(b) 
Sum of distance travelled in $t^{t h}$ and $(t+1)^{t h}$ seconds is $100 cm$
$u+\frac{a}{2}(2 t-1)+u+\frac{a}{2}(2(t+1)-1)=100$
$2 u+2 a t=100$
$u+a t=100 / 2$
$\Rightarrow v=50 cm / s$

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 $( t +1)^{ th }$ seconds is $100 cm$, then its velocity after $t$ seconds, in $.........cm / s$, is

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