Q6E - University Physics with Modern Physics

Question

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

Step-by-Step Solution

Answer

(a)The average velocity of the car between t=0 s to t=2.00 s is .

(b)The average velocity of the car between t=0 s to t=4.00 s is .

(c)The average velocity of the car between t=2.00 s to t=4.00 s is .

Step 1: Identification of the given data

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

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

Step 2:(a) Calculation of the average velocity between t = 0 s to t = 2.00 s

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 $v_{avg}^{velocity}=2.8m/s$

Step 3: (b) Calculation of the average velocitybetween t = 0 s to t = 4.00 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 $v_{avg}^{velocity}=5.2m/s$

Step 4: (c) Calculation of the average velocitybetween t = 2.00 s to t = 4.00 s

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

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

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,

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

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 $v_{avg}^{velocity}=7.6m/s$