A Honda Civic travels in a straight line along a road. The car’s distance x...

Step 1
a) The position of the car as a function of time t is given by
$x(t)=\alpha t^2-\beta t^3$
where
$$\begin{gathered} \alpha =1.50m/s^2 \\ \beta =0.05m/s^3 \end{gathered}$$
The average velocity is given by the ratio between the displacement and the time taken:
$v=\frac{\mathrm{\Delta }x}{\mathrm{\Delta }t}$
The position at t = 0 is:
$x(0)=\alpha \times 0^2-\beta \times 0^3=0$
The position at t = 2.00 s is:
$x(2)=\alpha \times 2^2-\beta \times 2^3=5.6m$
So the displacement is
$\mathrm{\Delta }x=x(2)-x(0)=5.6-0=5.6m$
The time interval is
$\mathrm{\Delta }t=2.0s-0s=2.0s$
And so, the average velocity in this interval is
$v=\frac{5.6m}{2.0s}=2.8m/s$
b) The position at t = 0 is:
$x(0)=\alpha \times 0^2-\beta \times 0^3=0$
While the position at t = 4.00 s is:
$x(4)=\alpha \times 4^2-\beta \times 4^3=20.8m$
So the displacement is
$\mathrm{\Delta }x=x(4)-x(0)=20.8-0=20.8m$
The time interval is
$\mathrm{\Delta }t=4.00-0=4.0s$
So the average velocity here is
$v=\frac{20.8}{4.0}=5.2m/s$
c) The position at t = 2 s is:
$x(2)=\alpha \times 2^2-\beta \times 2^3=5.6m$
While the position at t = 4 s is:

$x(4)=\alpha \times 4^2-\beta \times 4^3=20.8m$
So the displacement is
$\mathrm{\Delta }x=20.8-5.6=15.2m$
While the time interval is
$\mathrm{\Delta }t=4.0-2.0=2.0s$
So the average velocity is
$v=\frac{15.2}{2.0}=7.6m/s$