A car is stopped at a traffic light. It then travels along a straight road such that its distance from the light is given by$x\left(t\right)=b{t}^2-c{t}^3$, where$b=2.40m/s^2$and$c=0.120m/s^3$. (a) Calculate the average velocity of the car for the time interval t = 0 to t = 10.0 s. (b) Calculate the instantaneous velocity of the car at t = 0, t = 5.0 s, and t = 10.0 s. (c) How long after starting from rest is the car again at rest?

The distance function of the car is$x\left(t\right)=b{t}^2-c{t}^3$

using the values$b=2.40m/s^{2}andc=0.120m/s^3$

Therefore, the distance function will be,

$x\left(t\right)=\left(2.40m/s^2\right)\times t^2-\left(0.120m/s^3\right)\times t^3$

The final time is$t_f=10.0 \text{s}$$t_f=10.0 \text{s}$

The initial time is,localid="1655218991803" $t_i=0\text{s}$

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

The final position of the car is,

localid="1655219000319" $$\begin{gathered} x\left(t_f\right)=\left(2.40 {\text{m/s}}^{\text{2}}\right)\times {\left(10.0 \text{s}\right)}^2-\left(0.120 {\text{m/s}}^{\text{3}}\right)\times {\left(10.0 \text{s}\right)}^3 \\ =120\text{m} \end{gathered}$$

The initial position of the car is

localid="1655219012582" $$\begin{gathered} x\left(t_f\right)=\left(2.40 {\text{m/s}}^{\text{2}}\right)\times {\left(0 \text{s}\right)}^2-\left(0.120 {\text{m/s}}^{\text{3}}\right)\times {\left(0 \text{s}\right)}^3 \\ =0\text{m} \end{gathered}$$

Therefore, the average velocity can be expressed as,

localid="1655219019898" $v_{avg}^{velocity}=\frac{x\left(t_f\right)-x\left(t_i\right)}{t_f-t_i}$………………..(i)

Substituting values in the above expression,

$\begin{array}{rcl}{v}_{avg}^{velocity}& =& \frac{120m-0m}{10.00s-0s}\\ & =& 12m/s\end{array}$

The average velocity of the car between t = 0 s to t = 10.0 s is .

The expression of the instantaneous velocity can be determined by the first-order-derivative of the distance function can be given as,

$$\begin{gathered} v_1=\frac{ds}{dt} \\ =\frac{d}{dt}\left(\left(2.40m/s^2\right)\times t^2-\left(0.120m/s^2\right)\times t^2\right) \\ =\left(4.80m/s^2\right)\times t-\left(0.360m/s^2\right)\times t^2 \end{gathered}$$

For $t=0$,the instantaneous velocity will be,

$\begin{array}{rcl}{v}_0& =& \left(4.80m/s^2\right)\times \left(os\right)-\left(0.360m/s^3\right)\times {\left(s\right)}^3\\ & =& 0m/s\end{array}$

For,$t=5.0$the instantaneous velocity will be,

$\begin{array}{rcl}{v}_5& =& \left(4.80m/s\right)\times \left(5.0s\right)-\left(0.360m/s^2\right)\times {\left(5.0s\right)}^2\\ & =& 15m/s\end{array}$

Forrole="math" localid="1655223717357" $t=10.0s$, the instantaneous velocity will be,

The instantaneous velocity at t = 0 s, t = 5.0 s, and 10.0 s is, respectively.

$\begin{array}{rcl}{v}_{10.0}& =& \left(4.80m/s\right)\times \left(10.0s\right)-\left(0.360m/s^2\right)\times {\left(10.0s\right)}^2\\ & =& 12m/s\end{array}$

The instantaneous velocity at t = 0 s, t = 5.0 s, and 10.0 s is $0m/s,39m/s,12m/s$, respectively.

Substituting $t=10.0\text{s}$the distance function of the car

$\begin{array}{rcl}x\left(10.0\right)& =& \left(2.40 {\text{m/s}}^{\text{2}}\right)\times {\left(10.0 \text{s}\right)}^2-\left(0.120 {\text{m/s}}^{\text{3}}\right)\times {\left(10.0 \text{s}\right)}^3\\ & =& 120\text{m}\\ & & \end{array}$

Thus, the total distance covered by the car in t = 10.0 s is$120m$.