Question

A particle initially starts from rest, travels a distance Y in the first two seconds and a distance of X in next two seconds, if the body is moving with constant acceleration then :

• A. X = 2Y
• B. X + Y = 4X
• C. X + Y = 4Y
• D. X = 3Y

Answer 1

Answer: (C), (D)

The correct options are
C X + Y = 4Y
D X = 3Y

Let the acceleration of the particle be $a$.

Initial speed of the particle $u=0m/s$

Distance travelled by it in first 2 seconds, $Y=ut+\frac{1}{2}a{t}^2$ where $t=2$ s

$\therefore$ $Y=0t+\frac{1}{2}a(2{)}^2$

We get $Y=2a$ ....(1)

Velocity of the particle after $t=2$ s, $v=u+at$

$\therefore$ $v=0+a\left(2\right)=2a$

Distance covered by it next 2 seconds, $X=vt+\frac{1}{2}a{t}^2$ where $t=2$ s

$\therefore$ $X=\left(2a\right)2+\frac{1}{2}a(2{)}^2=6a$ .....(2)

From (1) and (2), we get $X=3Y$

Also $X+Y=3Y+Y=4Y$

Hence options C and D are correct.