Example 1: Physics Relationship

You are driving along in your car at 132 feet per second when you see a brick wall directly ahead.

When your car is 400 ft from the wall, you slam on the brakes, decelerating at a constant rate of $22\frac{\mathit{ft}}{s^2}$.

Do you stop before you hit the brick wall?

Step 1: Identify all your given data in terms of the physics relationships position , s(t) , velocity , v(t) , and acceleration , a(t) .

1)      “You are driving along in your car at 132 feet per second when you see a brick wall directly ahead.”

Your initial velocity is 132 feet per second, v(0) = 132 feet per second.

2)      When your car is 400 ft from the wall, you slam on the brakes, decelerating at a constant rate of $22\frac{\mathit{ft}}{s^2}$.

Your initial position would not be the 400 ft as that amount is not where you are starting, that is just letting you know how far you are from the wall. Instead, you would want your initial position to be zero since at time zero you have gone 0 distance, s(0) = 0 .

Your acceleration equation is the constant $22\frac{\mathit{ft}}{s^2}$, and since you are decelerating at that rate you would need the value to be negative , $a\left(t\right)=–22\frac{\mathit{ft}}{s^2}$.

Step 2: Setup and apply the Initial Value Indefinite Integral process.

In this problem you will need to find both the exact velocity equation and the exact position equation.

This is because you will need to use the velocity equation to determine when the car comes to a stop, (i.e., the velocity is zero ).

Then you will need to use the time it takes to stop in the position equation to see if you travel a distance less than or greater than the distance to the wall, 400ft .

First Initial Value Indefinite Integral Process : Getting velocity , v(t) ,from acceleration , a(t) .

$a\left(t\right)=–22\frac{\mathit{ft}}{s^2}$

$v\left(0\right)=132\frac{\mathit{ft}}{\mathit{sec}}$

$v\left(t\right)=\int a\left(t\right)\mathit{ dt}$

$v\left(t\right)=\int –22 \mathit{dt}$

Apply the Power Rule:

$v\left(t\right)=–22t\mathit{+}\mathit{C}$

Use the initial value, $v\left(0\right)=132\frac{\mathit{ft}}{\mathit{sec}}$ , to solve for the +C :

$v\left(0\right)=–22\left(0\right)\mathit{+}\mathit{C}\mathit{=}132$

$\mathit{C}\mathit{=}132$

Put it all together for the exact velocity equation, $v\left(t\right)$ .

$v\left(t\right)=–22t\mathit{+}132$

$v\left(t\right)=–22t\mathit{+}132$

Second Initial Value Indefinite Integral Process: Getting position , s(t) ,from velocity , v(t) .

$v\left(t\right)=–22t\mathit{+}132$

$s\left(0\right)=\mathit{0}\mathit{ }\mathit{ft}$

$s\left(t\right)=\int v\left(t\right)\mathit{ dt}$

$s\left(t\right)=\int –22t\mathit{+}132 \mathit{dt}$

Apply the Power Rule:

$s\left(t\right)=\frac{–22t^2}{2}+132t\mathit{+}\mathit{C}$

$s\left(t\right)=–11t^2+132t\mathit{+}\mathit{C}$

Use the initial value, $s\left(0\right)=\mathit{0}\mathit{ }\mathit{ft}$ , to solve for the +C :

$s\left(0\right)=–11{\left(0\right)}^2+132\left(0\right)\mathit{+}\mathit{C}=0$

$\mathit{C}=\mathit{0}$

Put it all together for the exact position equation, s(t) .

$s\left(t\right)=–11t^2+132t\mathit{+}\mathit{0}$

$s\left(t\right)=–11t^2+132t$

Step 3: Answer the actual question.

At this point you have all your physics equations ready to go. Now you just have a little algebra left to get to the actual answer to the question, will you hit the wall?

First, you need to figure out the amount of time the car will take to stop, or in physics language when will the velocity equal zero, v( t ) = 0 .

Second, you need to determine how far the car will have traveled based on the amount of time it takes the car to stop, s( t ) = ? .

When does the car stop, v( t ) = 0 ?

$v\left(t\right)=–22t\mathit{+}132$

$0=–22t\mathit{+}132$

$\begin{array}{c}0\\ +22t\end{array}=\begin{array}{c}–22t\\ +22t\end{array}\mathit{+}132$

$22t=132$

$\frac{22t}{22}=\frac{132}{22}$

$t=6\mathit{ seconds}$

What is the cars position when it stops, s( 6 ) = ?

$s\left(t\right)=–11t^2+132t$

$t=6\mathit{ seconds}$

$s\left(6 \right)=–11{\left(6 \right)}^2+132\left(6 \right)$

$s\left(6 \right)=396\mathit{ ft}$

Final Result:

It took the car 6 seconds to come to a stop, and during those 6 seconds the car traveled 396 ft.

You DID NOT hit the wall because the wall was 400ft in front of you and you only traveled 396 ft. You stopped 4ft before the wall.