Question

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 √ 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.

Answer 1

Answer:

$v(t)=-\frac{5}{2}\sqrt{t^3}+20\\s(t)=-\sqrt{t^5}+20t$

$s(t=4)=48\text{ m}$

Explanation:

In this case acceleration is defined as:

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

where k is a constant to be found.

To find the expressions for velocity and position as a function of time you must integrate the expression above for acceleration two times.

Initial conditions and boundary conditions are defined with the rest of the data as:

$v(t=0)=20\text{ m/s}\\v(t=4)=0\text{ m/s}\\s(t=0)=0\text{ m}$

First integration is equal to:

$a'(t)=v(t)=-k\int\sqrt{t}dt=-\frac{2}{3}k\sqrt{t^3}+C_1$

The boundary condition and initial condition can be used to calculate $k$ and $C_1$:

$C_1=20\\k=\frac{15}{4}$

With this expression for velocity is defined as:

$v(t)=-\frac{5}{2}\sqrt{t^3}+20$

The same can be done to get to expression for position:

$s(t)=-\sqrt{t^5}+20$

To get the total distance traveled you can integrate the velocity expression from time=0 sec to time=4 sec:

$s_{tot}=\int_0^4(-\frac{5}{2}\sqrt{t^3}+20)dt=48\text{ m}$