Question

A particle moves rectilinearly possessing a parabolic $s-t$ graph. Find the average velocity of the particle over a time interval from $t=\frac{1}{2}s$ to $t=1.5s$.

Answer 1

First we know the equation of motion of the particle will be $s=at-b{t}^2$

Now, from the figure we know that parabola is symmetric above $t=1$

Or, $\frac{ds}{dt}=0,at$ $t=1$

so we get, $a-2bt=0$ or $a=2b........\left(1\right)$

Using (1)

$s_{0.5}=a\left(0.5\right)-b(0.5{)}^2=\frac{3a}{8}$ and $s_{1.5}=a\left(1.5\right)-b(1.5{)}^2=\frac{3a}{8}$

So there is no displacement between $t=0.5$ and $t=1.5$

hence the $v_{avg}=\frac{displacement}{time}$ $=\frac{0}{1.5-0.5}=0$