Chapter 11.1, Problem 11.12P

Many car companies are performing research on collision avoidance systems. A small prototype applies engine braking that decelerates the vehicle according to the relationship $a=-k^{\sqrt{t}}$ , where a and t are expressed in m/s² and seconds, respectively. The vehicle is traveling at 20 m/s when its radar sensors detect a stationary obstacle. Knowing that it takes the prototype vehicle 4 seconds to stop, determine (a) expressions for its velocity and position as a function of time, (b) how far the vehicle traveled before it stopped.

To determine

(a)

The expression for the velocity as a function of time.

The expression for the position as a function of time.

Answer to Problem 11.12P

The expression for the velocity as a function of time is $v=-\frac{2}{3}k{t}^{3/2}+20\text{m}/\text{s}$.

The expression for the position as a function of time is $x\left(t\right)=-\frac{4}{15}k\left[{\left(t\right)}^{5/2}\right]+\left(20\text{m}/\text{s}\right)\left(t\right)$.

Explanation of Solution

Given information:

The initial velocity is $20\text{m}/\text{s}$, the time take to stop the prototype vehicle is $\text{4}\text{s}$ and the expression for the deceleration is $a=-k\sqrt{t}$.

Write the expression for the deceleration.

$a=-k\sqrt{t}$ ...... (I)

Here, the time is $t$ and the constant is $k$.

Write the expression for the velocity.

$v=\int adt$ ….. (II)

Substitute $-k\sqrt{t}$ for $a$ in Equation (II).

$\begin{array}{c}v=\int \left(-k\sqrt{t}\right)dt\\ =-k\int \left(\sqrt{t}\right)dt\\ =-k\left[\frac{t^{1/2+1}}{\frac{1}{2+1}}\right]+C\\ =-\frac{2}{3}k{t}^{3/2}+C\end{array}$ ...... (III)

Here, the integration constant is $C$.

Write the expression for the position.

$x\left(t\right)=\int vdt$ ...... (IV)

Calculation:

Substitute $20\text{m}/\text{s}$ for $v$ and $0$ for $t$ in Equation (III).

$\begin{array}{l}20\text{m}/\text{s}=-\frac{2}{3}k{\left(0\right)}^{3/2}+C\\ 20\text{m}/\text{s}=0+C\\ C=20\text{m}/\text{s}\end{array}$

Substitute $20\text{m}/\text{s}$ for $C$ in Equation (III).

$v=-\frac{2}{3}k{t}^{3/2}+20\text{m}/\text{s}$ ...... (V)

Substitute $-\frac{2}{3}k{t}^{3/2}+20\text{m}/\text{s}$ for $v$ in Equation (IV).

$\begin{array}{c}x\left(t\right)=\int \left(-\frac{2}{3}k{t}^{3/2}+20\text{m}/\text{s}\right)dt\\ =-\frac{2}{3}k\left[\frac{t^{3/2+1}}{\frac{3}{2}+1}\right]+\left(20\text{m}/\text{s}\right)t+C\\ =-\frac{4}{15}k\left[t^{5/2}\right]+\left(20\text{m}/\text{s}\right)t+C\end{array}$ ...... (VI)

Substitute $0$ for $t$ in Equation (V).

$\begin{array}{l}x\left(0\right)=-\frac{4}{15}k\left[{\left(0\right)}^{5/2}\right]+\left(20\text{m}/\text{s}\right)\left(0\right)+C\\ 0=0+0+C\\ C=0\end{array}$

Substitute $0$ for $C$ in Equation (V).

$\begin{array}{l}x\left(t\right)=-\frac{4}{15}k\left[{\left(t\right)}^{5/2}\right]+\left(20\text{m}/\text{s}\right)\left(t\right)+0\\ x\left(t\right)=-\frac{4}{15}k\left[{\left(t\right)}^{5/2}\right]+\left(20\text{m}/\text{s}\right)\left(t\right)\end{array}$

Conclusion:

The expression for the velocity as a function of time is $v=-\frac{2}{3}k{t}^{3/2}+20\text{m}/\text{s}$.

The expression for the position as a function of time is $x\left(t\right)=-\frac{4}{15}k\left[{\left(t\right)}^{5/2}\right]+\left(20\text{m}/\text{s}\right)\left(t\right)$.

To determine

(b)

The distance travelled by the vehicle before stop.

Answer to Problem 11.12P

The distance travelled by the vehicle before stop is $48\text{m}$.

Explanation of Solution

Given information:

The initial velocity is $20\text{m}/\text{s}$, the time take to stop the prototype vehicle is $\text{4}\text{s}$ and the expression for the deceleration is $a=-k\sqrt{t}$.

Write the expression for the distance travelled by the vehicle.

$x\left(t\right)=-\frac{4}{15}k\left[{\left(t\right)}^{5/2}\right]+\left(20\text{m}/\text{s}\right)\left(t\right)$ ...... (VII)

Write the expression for the velocity.

$v=-\frac{2}{3}k{t}^{3/2}+20\text{m}/\text{s}$ ...... (VIII).

Calculation:

Substitute $4\text{s}$ for $t$ and $0$ for $v$ in Equation (VIII).

$\begin{array}{l}0=-\frac{2}{3}k{\left(4\text{s}\right)}^{3/2}+20\text{m}/\text{s}\\ -k\left(8{\text{s}}^{3/2}\right)=\frac{3}{2}\left(-20\text{m}/\text{s}\right)\\ k=\frac{30\text{m}/\text{s}}{8{\text{s}}^{3/2}}\\ k=3.75\text{m}/{\text{s}}^{5/2}\end{array}$

Substitute $4\text{s}$ for $t$ and $3.75\text{m}/{\text{s}}^{5/2}$ for $k$ in Equation (VII).

$\begin{array}{c}x\left(4\text{s}\right)=-\frac{4}{15}\left(3.75\text{m}/{\text{s}}^{5/2}\right)\left[{\left(4\text{s}\right)}^{5/2}\right]+\left(20\text{m}/\text{s}\right)\left(4\text{s}\right)\\ =-\frac{4}{15}\left(3.75\text{m}/{\text{s}}^{5/2}\right)\left[\left(32{\text{s}}^{5/2}\right)\right]+\left(80\text{m}\right)\\ =-32\text{m}+\left(80\text{m}\right)\\ =48\text{m}\end{array}$

Conclusion:

The distance travelled by the vehicle before stop is $48\text{m}$.