Question

A particle is moving in a straight line and passes through a point O with a velocity of $6\text{m/s}$. Then particle moves with a constant retardation of $2{\text{m/s}}^2$ for $4s$ and thereafter moves with constant velocity. How long after leaving O does the particle return to O?

• A. $3$ seconds
• B. $8$ seconds
• C. $8.5$ seconds
• D. $4.5$ seconds

Answer 1

The correct option is B $8$ seconds

Let the particle moves towards right with velocity 6 $m/s$. Due to retardation after time $t_1$, its velocity becomes zero.

From $v=u-at\Rightarrow 0=6-2\times t_1\Rightarrow \times t_1=3s$.

But, it retards for $4s$. It means, after reaching point $A$ direction of motion gets reversed and it accelerates for next one second.
$S_{OA}=u{t}_1-\frac{1}{2}a{t}_1^2=6\times 3-\frac{1}{2}\left(2\right)\times (3{)}^2$
$=18-9=9m$

$S_{AB}=\frac{1}{2}\times \left(2\right)\times (1{)}^1=1m$
$\therefore S_{BC}=S_{OA}-S_{AB}=9-1=8m$

Now, velocity of the particle at point $B$ in return journey:
$v=0+2\times 1=2\text{m/s}$
In return journey from $B$ to $C$ particle moves with constan velocity of 2 $\text{m/s}$ to cover the distance $8m$.

So, time taken is
$\frac{Distance}{Velocity}=\frac{8}{2}=4seconds$
Total time taken by particle to return at point O is
$T=t_{OA}+t_{AB}+t_{BC}T=3+1+4=8s$