Question

A car starts moving along a line, first with an acceleration $a=5m{s}^{-2}$ starting from rest, then uniformly and finally decelerating at the same rate, comes to rest in the total time of $25seconds\left(t_1\right)$, then average velocity during the time is equal to $v=72kmph$. How long does the particle move uniformly?

• A. $5seconds$
• B. $2.5hours$
• C. $1.5hours$
• D. $15seconds$

Answer 1

The correct option is A $5seconds$

Let the initial velocity$u=0$

The time taken and the distance traveled during accelerated motion and retarding motion will be same as the acceleration is same

For accelerated motion

$v=u+at=5t$

And for retardation

$0=v-5t$

$v=5t$

As the velocity v will become u for retardation

Total distance traveled

$=$ area of $v-t$

$=$ area$I+$ area $II+$ area $III$

$2\left(\frac{1}{2}5t^2\right)+5tx=$distance traveled during motion

$x=$time during which particle remains in uniform motion

$5t^2+5tx=$speed$x$ time

$5t^2+5tx=20\times 25$

$t^2+tx=100\to 1$

Also the total time is given $25$seconds

Thus $2t+x=25$

Thus $x=25-2t\to 2$

Substituting $2$ in $1$

$t^2+t(25-2t)=100$

$t^2-25t+100=0$

$t=5$ or $t=20$

Thus the time will be $t=5$sec as $t=20$s is not possible as it will exceed the given time

The particle remains in motion for $5$ seconds