Question

A particle is moving in a straight line with initial velocity u and uniform acceleration a .If the sum of the distances travelled in t^th and (t + 1)^th seconds is 100cm, then its velocity after t seconds is

Answer 1

If. A particle is moving in a straight line with initial velocity `u` and uniform acceleration `f`. If the sum of the distances travelled in `t^(th) and (t + 1)^(th)` seconds is `100 cm`, then its velocity after `t` seconds, in `cm//s

Answer 2

Given

A particle is moving in a straight line with initial velocity u and uniform acceleration a . The sum of the distances travelled in "t" th and "(t + 1)" th seconds is 100 cm

To Find

Velocity after t seconds

Formula Applied

$\bf \pink{\bigstar\ \; S_n=u+\dfrac{a}{2}[2n-1]}$

where ,

• Sₙ denotes displacement in nth second
• u denotes initial velocity
• a denotes acceleration
• n denotes nth second

Solution

Sum of the distances travelled in 't' th and (t+1) th  seconds is 100 cm

$\to \bf S_t+S_{t+1}=100\ cm\\\\ \to \rm \left\{ u+\dfrac{a}{2}(2t-1) \right\}+\left\{ u+\dfrac{a}{2}(2(t+1)-1) \right\}=100$

$\to \rm 2u+\dfrac{a}{2}(2t-1+2(t+1)-1)=100\\\\\to \rm 2u+\dfrac{a}{2}(2t-1+2t+2-1)=100\\\\\to \rm 2u+\dfrac{a}{2}(4t+2-2)=100\\\\\to \rm 2u+\dfrac{a}{2}(4t)=100\\\\\to \rm 2u+2at=100\\\\\to \rm 2(u+at)=100\\\\\leadsto \bf \blue{u+at=50\ \; \bigstar}$

Velocity after 't' seconds is given by ,

$\rm v=u+at\\\\\leadsto \bf \red{v=50\ cm/s\ \; \bigstar}$