Question

Many drivers' education books provide tables that relate a car's braking distance to the speed of the car. Utilize what you have learned about the stopping distance-velocity relationship tocomplete the table.

Answer 1

To solve this problem we must apply Torricelli equation:

`v^2=v^2_o+2*\alpha* d^{}_{}`

where:

v = final velocity

vo = initial velocity

α = acceleration

d = distance

Lets use the information of the first row of the table to find a:

`\begin{gathered} v^2=v^2_o+2*\alpha* d^{}_{} \\ 0^2=60^2+2*\alpha*240 \\ 0=3,600+480*\alpha \\ -480*\alpha=3,600 \\ \alpha=(3,600)/(-480) \\ \alpha=-7.5ft/s^2 \end{gathered}`

Now we can calculate distances a, b and c:

`\begin{gathered} v^2=v^2_{^{}o^{}}+2*\alpha* d \\ 0^2=40^2+2*(-7.5)* d \\ 0=1,600-15* d \\ 15* d=1,600 \\ d=(1,600)/(15) \\ d\approx107ft \end{gathered}`

So distance a = 107 ft.

Now we have:

`\begin{gathered} v^2=v^2_o+2*\alpha* d \\ 0^2=30^2+2^{}*(-7.5)* d \\ 0=900-15* d \\ 15* d=900 \\ d=(900)/(15) \\ d=60ft \end{gathered}`

So distance b = 60 ft.

Finally:

`\begin{gathered} v^2=v^2_o+2*\alpha* d \\ 0^2=10^2+2*(-7.5)* d \\ 0=100-15* d \\ 15* d=100 \\ d=(100)/(15) \\ d\approx7ft \end{gathered}`

So distance c = 7 ft.