Question

A particle experience constant acceleration for 20 second after starting from rest. If it travels a distance X in first 10 sec. and Y in next 10 sec. Find relation between X & Y?

Answer 1

$S=ut+\frac{1}{2}at^2$

First $10$ seconds :
At $t=0$, the velocity of particle is $0$ as it is starting from rest. Therefore, $u=0$ and the distance travelled is given by 
$X=0*10+\frac{1}{2}a*10^2=50a$

Next $10$ seconds :
By the end of $10$th second, the particle's velocity reaches $10a$. Therefore, $u=10a$ and the distance travelled is given by
$Y=10a*10+\frac{1}{2}a*10^2=150a$

Dividing gives
$\frac{Y}{X}=\frac{150a}{50a}=3$

$\implies\boxed{Y=3X}$