Question

A particle moves in a straight line and passes through $O$, a fixed on the line, with a velocity of $6m{s}^{-1}$. The particle moves with a constant retardation of $2m{s}^{-2}$ for $4$ seconds and thereafter moves with constant velocity. How long after leaving $O$ does particle return to $O$?

• A. $8s$
• B. $2s$
• C. $4s$
• D. $6s$

Answer 1

Answer: (B)

$2s$

acceleration , $a=-4m/s²$

Intial velocity , $u=8m/s$

Here, acceleration is negative sign which indicates particle is retarding motion.

Let after t time particle will be rest.

∴ final velocity , $v=0m/s$

Now, use formula , $v=u+at$

$0=8-4t$

$t=2sec$

hence, velocity of particle is decreases at $t=2sec$ after that particle moves in direction of Acceleration .

so, distance covered in $6sec$= distance travelled in first $2sec$ + distance travelled by last $4sec$

$=|ut₁+1/2at₁²|+|0.t₂+1/2at₂²|$

In last $4sec$ , initial velocity will take zero. Because particle will be rest after two sec.

Here, $t₁=2$ and $t₂=4$

Now, Total distance travelled $=8\times 2+1/2\times (-4)\times 2²+|1/2(-4)\times \left(4\right)²|$

$=16-8+|-32|$

$=8+32=40m$

Similarly , total distance travelled in $5sec$ = distance travelled in $2sec$ + distance travelled in last $3sec$

$=16-8+|-18|$

$=8+18=26m$