Chapter 3, Problem 9P

To determine

(a)

The distance traveled by the vehicle when the acceleration is 45 ft/sec.

Answer to Problem 9P

  $x=443\text{ft}$

Explanation of Solution

Given information:

  $\begin{array}{c}\frac{du}{dt}=3.6-0.06u\\ \alpha =3.6\\ \beta =0.06\end{array}$

Initial vehicle speed, $u_0=30\text{ft/sec}$

Final vehicle speed, $u_t=45\text{ft/sec}$

Concept used:

The time required by the vehicle upon acceleration is calculated by using the relation,

  $\begin{array}{c}-\beta t=\ln \left[\frac{\left(\alpha -\beta u_t\right)}{\left(\alpha -\beta u_0\right)}\right]\\ t=\left(-\frac{1}{\beta }\right)\ln \left[\frac{\left(\alpha -\beta u_t\right)}{\left(\alpha -\beta u_0\right)}\right]\end{array}$

The distance travelled at any time is calculated by using the formula,

  $x=\left(\frac{\alpha }{\beta }\right)t-\frac{\alpha }{{\beta }^2}\left(1-e^{-\beta t}\right)+\frac{u_0}{\beta }\left(1-e^{-\beta t}\right)$

Calculation:

The time taken by the vehicle is calculated as,

  $\begin{array}{c}t=\left(-\frac{1}{\beta }\right)\ln \left[\frac{\left(\alpha -\beta u_t\right)}{\left(\alpha -\beta u_0\right)}\right]\\ =\left(-\frac{1}{0.06}\right)\ln \left[\frac{3.6-0.06\left(45\right)}{3.6-0.06\left(30\right)}\right]\\ =11.55\sec \end{array}$

  $\begin{array}{c}x=\left(\frac{\alpha }{\beta }\right)t-\frac{\alpha }{{\beta }^2}\left(1-e^{-\beta t}\right)+\frac{u_0}{\beta }\left(1-e^{-\beta t}\right)\\ =\left(\frac{3.6}{0.06}\right)11.55-\frac{3.6}{{\left(0.06\right)}^2}\left(1-e^{-\left(0.06\times 11.55\right)}\right)+\frac{30}{0.06}\left(1-e^{-\left(0.06\times 11.55\right)}\right)\\ =693-499.93+249.96\\ =443\text{ft}\end{array}$

Conclusion:

The distance traveled when the vehicle has accelerated to 45 ft/sec is $443\text{ft}$.

To determine

(b)

The time taken by the vehicle to attain a speed of 45 ft/sec.

Answer to Problem 9P

  $t=11.55\sec$

Explanation of Solution

Given information:

  $\begin{array}{c}\frac{du}{dt}=3.6-0.06u\\ \alpha =3.6\\ \beta =0.06\end{array}$

Initial vehicle speed, $u_0=30\text{ft/sec}$

Final vehicle speed, $u_t=45\text{ft/sec}$

Concept used:

The time required by the vehicle after accelerated is calculated by using the relation,

  $\begin{array}{c}-\beta t=\ln \left[\frac{\left(\alpha -\beta u_t\right)}{\left(\alpha -\beta u_0\right)}\right]\\ t=\left(-\frac{1}{\beta }\right)\ln \left[\frac{\left(\alpha -\beta u_t\right)}{\left(\alpha -\beta u_0\right)}\right]\end{array}$

Calculation:

  $\begin{array}{c}t=\left(-\frac{1}{\beta }\right)\ln \left[\frac{\left(\alpha -\beta u_t\right)}{\left(\alpha -\beta u_0\right)}\right]\\ =\left(-\frac{1}{0.06}\right)\ln \left[\frac{3.6-0.06\left(45\right)}{3.6-0.06\left(30\right)}\right]\\ =11.55\sec \end{array}$

Conclusion:

The time for vehicle to attain speed of 45 ft/sec is $11.55\sec$.

To determine

(c)

The acceleration of the vehicle after 4 seconds.

Answer to Problem 9P

  $a=1.42\frac{\text{ft}}{{\text{s}}^{\text{2}}}$

Explanation of Solution

Given information:

  $\begin{array}{c}\frac{du}{dt}=3.6-0.06u\\ \alpha =3.6\\ \beta =0.06\end{array}$

Initial vehicle speed, $u_0=30\text{ft/sec}$

Final vehicle speed, $u_t=45\text{ft/sec}$

Time taken, $t=4\sec$

Concept used:

The initial velocity is calculated and then the acceleration is calculated.

  $u_t=\frac{\alpha }{\beta }\left(1-e^{-\beta t}\right)+u_0{e}^{-\beta t}$

  $a=\frac{du}{dt}$

Calculation:

  $\begin{array}{c}{u}_t=\frac{\alpha }{\beta }\left(1-e^{-\beta t}\right)+u_0{e}^{-\beta t}\\ =\frac{3.6}{0.06}\left(1-e^{-\left(0.06\times 4\right)}\right)+30\left(e^{-\left(0.06\times 4\right)}\right)\\ =\frac{3.6}{0.06}\left(1-0.78663\right)+30\left(0.78663\right)\\ =36.40\text{ft/sec}\end{array}$

  $\begin{array}{c}a=\frac{du}{dt}=3.6-0.06u\\ =3.6-0.06\left(36.40\right)\\ =1.42\frac{\text{ft}}{{\text{s}}^{\text{2}}}\end{array}$

Conclusion:

The acceleration after 4 seconds is $1.42\frac{\text{ft}}{{\text{s}}^{\text{2}}}$.