Question

The position of an object moving along x-axis is given by $x=a+b{t}^2$

where $a=8.5m,b=2.5m{s}^2$ and $t$ measured in seconds. What is its velocity at $t=0s$ and $t=2.0s$. What is the velocity between $t=2.0s$ and $t=4.0s$?

Answer 1

$x\left(4\right)=a+b\times 4^2=16b$

$x\left(2\right)=a+b\times 2^2=4b$

$x\left(4\right)-x\left(2\right)=16b-4b=12b$

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

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

At t = 0, v = 0

t = 2, v = 4b = 4 x 2.5 = 10 m/s

Velocity between 2 & 4 $=\frac{x\left(4\right)-x\left(2\right)}{4-2}$

$=\frac{b[16-4]}{2}=6b=15$ m/s