Question

A car accelerates from rest at constant rate for first $10s$ and covers a distance $x$ . It covers a distance $y$ in next $10s$ at the same acceleration. Which of the following is true?

• A. $x=3y$
• B. $y=3x$
• C. $x=y$
• D. $y=2x$

Answer 1

Answer: (B)

From equation of motion, we have
$s=ut+\frac{1}{2}a{t}^2$
where $u$ is initial velocity, $t$ is time, $a$ is acceleration.
Since car accelerates from rest $u=0,t=10s$
$\therefore s=0+\frac{1}{2}\times a\times (10{)}^2=50a\dots$ (i)
Also, $v=u+at$
where, $v$ is final velocity.
$\therefore$ Velocity after $10s$ is
$v=0+a\times 10$
$v=10a=10\times \frac{s}{50}\dots$ (ii)
In the next $10s$ car moves with constant acceleration and with initial velocity
$\therefore s^{\prime }=vt+\frac{1}{2}a{t}^2$
$=\frac{s}{50}\times 10\times 10+\frac{1}{2}\times \frac{s}{50}\times 100=3s$
Given, $s=x$ and $s^{\prime }=y$
$\therefore y=3x$