JEE Physics 3

Let $Vi$ be the initial velocity and $a$ be the acceleration, $Then$ $S=Vi*t + 0.5*a*t^2$ , We have $S=200$ and $t=2,$ $200=2Vi +2a,$

Vi+a=100

Velocity at end of 2 seconds = $Vi +(100-Vi)*2$

$V(2)=200-Vi$

For next 4 seconds body traveled $220 cm$ ,

We have initial V at the start of the next phase= $V(2)$

and $t=4$ and $a=100-Vi$

Plugging this into the equation,

$(200-Vi)*4 +(100-Vi)*0.5*4*4=220$

$1600-12Vi=220.$

$Vi=115$

$a=100-115=-15 cm/s^2$

Velocity at end of 7 seconds = $V(7)= Vi+a*7$

$V(7)=115-15*7=115-105=10 cm/s.$