A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by the equation$x\left(t\right)=\alpha t^2-\beta t^3$ where role="math" localid="1655226337795" $\alpha =1.50 m/s^2$androle="math" localid="1655226362269" $\beta =0.0500m/s^2$ . Calculate the average velocity of the car for each time interval: (a) t = 0 to t = 2.00 s; (b) t = 0 to t = 4.00 s; (c) t = 2.00 s to t = 4.00 s.

The distance function of the car is$x\left(t\right)=\alpha t^2-\beta t^3$

$\alpha =1.50m/s^2,and\beta =0.0500m/s^2$

Therefore the distance function of the car will be

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

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,

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

The initial position of the car is,

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

Therefore, the average velocity can be expressed as,

$v_{avg}^{velocity}=\frac{x\left(t_f\right)-x\left(t_i\right)}{t_f-t_i}$………………..(i)

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,

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

The initial position of the car is,

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

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, $t_i=2.00\text{s}$

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

The initial position of the car is,

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

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$v_{avg}^{velocity}=7.6m/s$