Question

The position of a particle along along x-axis is given by $x=a+b{t}^2$, where a= 8.5 m, b = 2.5 m and $t$ is in seconds. What is its average velocity at t = 2.0 s? What will be its average velocity between $t=2s$ & $t=4s$?

Answer 1

Given that,

$a=8.5m$

$b=2.5m$

$t=2\sec$

$x=a+b{t}^2$

The velocity is

$v=\frac{dx}{dt}=2bt$

Now, velocity at t = 0 is 0

Velocity at t = 2s is

$v=2\times 2.5\times 2$

$v=10m/s$

Now, the distance at 2s

$x=8.5+2.5\times 4$

$x=18.5m$

Now, the distance at 4 s

$x=8.5+2.5\times 16$

$x=48.5m$

Now, average velocity is

$v=\frac{dis\tan ce}{time}$

$v=\frac{x_2-x_1}{2}$

$v=\frac{(48.5-18.5)}{2}$

$v=\frac{30}{2}$

$v=15m/s$

Hence, the average velocity is 15 m/s