91Ó°ÊÓ

The distance function of the car is$ x\left(t\right)=\mathrm{Î\pm }{t}^2-\mathrm{β}{t}^3$

$\mathrm{Î\pm }=1.50m/s^2,and\mathrm{β}=0.0500m/s^2$

Therefore the distance function of the car will be

$x\left(t\right)=\left(1.50  {\text{m/s}}^2\right)×t^2-\left(0.0500 {\text{m/s}}^2\right)×t^3$

The final time is, $t_f=2.00 \text{s}$

The initial time is, $t_i=0\text{s}$

Substituting these values in the distance function of the car gives,

The final position of the car is,

$\begin{array}{rcl}x\left(t_f\right)& =& \left(1.50  {\text{m/s}}^2\right)×(2.00\text{s}{)}^2-\left(0.0500 {\text{m/s}}^2\right)×(2.00\text{s}{)}^3\\ & =& 5.6\text{m}\\ & & \end{array}$

The initial position of the car is,

$\begin{array}{rcl}x\left(t_i\right)& =& \left(1.50  {\text{m/s}}^2\right)×(0\text{s}{)}^2-\left(0.0500 {\text{m/s}}^2\right)×(0\text{s}{)}^3\\ & =& 0\text{m}\\ & & \end{array}$

Therefore, the average velocity can be expressed as,

$v_{avg}^{velocity}=\frac{x\left(t_f\right)-x\left(t_i\right)}{t_f-t_i}$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦..(¾±)

Substituting values in the equation (i), gives

$\begin{array}{rcl}{v}_{avg}^{velocity}& =& \frac{5.6\text{m}-0\text{m}}{2.00\text{s}-0\text{s}}\\ & =& 2.8m/s\end{array}$

Thus, the average velocity of the car is$v_{avg}^{velocity}=2.8m/s$

The final time is,$t_f=4.00 \text{s}$

The initial time is, $t_i=0\text{s}$

The final position of the car is,

$\begin{array}{rcl}x\left(t_f\right)& =& \left(1.50  {\text{m/s}}^2\right)×(4.00\text{s}{)}^2-\left(0.0500 {\text{m/s}}^2\right)×(4.00\text{s}{)}^3\\ & =& 20.8\text{m}\\ & & \end{array}$

The initial position of the car is,

$\begin{array}{rcl}x\left(t_i\right)& =& \left(1.50  {\text{m/s}}^2\right)×(0\text{s}{)}^2-\left(0.0500 {\text{m/s}}^2\right)×(0\text{s}{)}^3\\ & =& 0\text{m}\\ & & \end{array}$

Substituting these values in the equation (i) gives

$\begin{array}{rcl}{v}_{avg}^{velocity}& =& \frac{20.8\text{m}-0\text{m}}{4.00\text{s}-0\text{s}}\\ & =& 5.2m/s\end{array}$

Thus, the average velocity of the car is$v_{avg}^{velocity}=5.2m/s$

The final time is, $t_f=4.00 \text{s}$

The initial time is, $t_i=2.00\text{s}$

$\begin{array}{rcl}x\left(t_f\right)& =& \left(1.50  {\text{m/s}}^2\right)×(4.00\text{s}{)}^2-\left(0.0500 {\text{m/s}}^2\right)×(4.00\text{s}{)}^3\\ & =& 20.8\text{m}\\ & & \end{array}$

The initial position of the car is,

$\begin{array}{rcl}x\left(t_i\right)& =& \left(1.50  {\text{m/s}}^2\right)×(2.00\text{s}{)}^2-\left(0.0500 {\text{m/s}}^2\right)×(2.00\text{s}{)}^3\\ & =& 5.6\text{m}\\ & & \end{array}$

Substituting these values in the equation (i) gives

$\begin{array}{rcl}{v}_{avg}^{velocity}& =& \frac{20.8\text{m}-5.6\text{m}}{4.00\text{s}-2.00\text{s}}\\ & =& 7.6m/s\end{array}$

Thus, the average velocity of the car is$v_{avg}^{velocity}=7.6m/s$