Q7E - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

(a)The average velocity of the car between t = 0 s to t = 10.0 s is $12m/s$.

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

(c) Total distance covered by the car in t = 10.0 s is$120m$.

 

Step 1: Identification of the given data

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

using the values$b=2.40m/s^2 and c=0.120m/s^3$

 Therefore, the distance function will be, 

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

Step 2: (a) Calculation of the average velocity between t = 0 s to t = 10.0 s

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

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

The initial position of the car is

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

Therefore, the average velocity can be expressed as,

 

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

Step 3: (a) Calculation of theinstantaneous velocity at t = 0 s, t = 5.0 s, and t = 10.0 s

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,

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

 For $t=10.0s$ , 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.

Step 4: (c) Calculation ofthe total distance

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$120m$.