 eiraszero11cu

2022-01-07

A car braked with a constant deceleration of 16 ft/s^2 , producing skid marks measuring 200 ft before coming to a stop. How fast was the car traveling when the brakes were first applied? Travis Hicks

Step 1
$a\left(t\right)=-16\frac{ft}{{s}^{2}}$
We can make it the forward direction is positive, since acceleration is in the opposite direction of forvard if is slowing down.
Step 2
Find the antiderivative, which is velocity
$v\left(t\right)=-16t+{v}_{0}$
At this point ${v}_{0}$ is the unknown constant, and also what the problem is asking for, the initial speed.t=0 is the time when the brakes are first applied.
Step 3
Find the antiderivative, which is the position
$s\left(t\right)=\frac{-16}{2}{t}^{2}+{v}_{0}t+C$
If we let the position at time t=0 be the 0 position, then C=0
$s\left(t\right)=-8{t}^{2}+{v}_{0}t$ Stella Calderon

Step 4
When the position is 200 ft, we have
$s\left(t\right)=200=-8{t}^{2}+{v}_{0}t$
$8{t}^{2}+200={v}_{0}t$
$\frac{8{t}^{2}+200}{t}={v}_{0}$
Step 5
When the velosity is 0 we have
$v\left(t\right)=0=-16t+{v}_{0}$
$16t={v}_{0}$ karton

Step 6
Set the two equal to each other and solve for t.
$\begin{array}{}\frac{8{t}^{2}+200}{t}=16t\\ 8{t}^{2}+200=16{t}^{2}\\ 200=8{t}^{2}\\ {t}^{2}=25\\ t=±5\end{array}$
We use the positive number, t=5 s
(the time it took to come to a stop)
Step 7
Plug into one of the expressions we have for ${v}_{0}$
${v}_{0}=16\left(5\right)=80ft/s$
Result
80 ft/s

Do you have a similar question?