# Speed of Impact Typical

1. Jan 22, 2010

### IronBrain

1. The problem statement, all variables and given/known data

I've seen this problem on here only twice, once never answered, and the other just saying the OP's "solution" was on the right track

Problem:

Code (Text):

You are arguing over a cell phone while trailing an unmarked police car by 37 m. Both your car and the police car are travelling at 110 km/h. Your argument diverts your attention from the police car for 1.5 s (long enough for you to look at the phone and yell, "I won't do that!"). At the beginning of that 1.5 s, the police officer begins emergency braking at 5 m/s2.

Question
Code (Text):

B.
(b) Suppose that you take another 0.4 s to realize your danger and begin braking. If you too brake at 5 m/s2, what is your speed when you hit the police car?

I've ran out of attempts to earn points for this equation however, I want to know how to do it correctly, I looked at a previous example somewhere, best I can find.

2. Relevant equations

Question A:
Code (Text):

(a) What is the separation between the two cars when your attention finally returns?

I solved this by simply converting the speed into m/s finding the amount of distance the cop travelled when he began to de accelerate for 1.5's at -5 m/s^2 and then the distance at which I travel for 1.5's, Adding the distance travelled for duration to 37 m (position of cop car) and adding the distance which I travel for duration to 0 to assume something similar to x/y plane, I arrived at 32 m for that duration which was correct.

3. The attempt at a solution
I've ran out of attempts to earn points for this equation however, I want to know how to do it correctly, I looked at a previous example somewhere, best I can find.

First going back to square 1, I have a new time t = 1.9, I use the related question above formula to find the distance travelled by both cop car and my car for this duration. As shown here

Formula to find my distance travelled

$Dist = Speed x Time = 30.55 m/s x 1.9 = 58.045 m$

Distance travelled by cop car de-accelerating

$D(t) = -2.5t^2+30.55t = -2.5(1.9^2)+30.55(1.9) = 49.02$

To find new distance of separation, adding the distance travel when cop begins to brake to 37, equates to 86.02 subtracting my distance from this distance equates to 86.02 - 58.045 = 27.975

Now after this time duration I begin to de-accelerate as well, I first find the velocity of the cop at the end of this time period, here, equates to

$v(t) = -5(1.9)+30.55 = 21.05$

Making a new equations for both car's positions and new equation for cop's velocity so I can find the time which they impact or "intersect" I arrive to

Cop's New Velocity Formula
$-5(t)+21.05$

Cop's Position Formula
$-2.5(t^2)+21.05(t)+27.975$

My Car's Velocity Formula
$-5(t) + 30.55$

My Car's Position Formula
$-2.5(t^2)+30.55(t)$

Now setting these two new position formulas/functions to each other and solving for t, and plugging the time into my car's velocity formula and converting the units over back to km/h I will arrive at a possible solution

Solving for time, t.

$-2.5(t^2)+30.55(t) = -2.5(t^2)+21.05(t)+27.975$

$30.55(t)-21.05(t) = 27.975$

$9.5t=27.975$

$t = 2.9394 s$

Plugging this into the formula for my velocity

$-5(2.9394)+30.55 = 15.853 m/s$

Converting this back to km / h equates to 57.0708 km/h

Is this correct? The only mistake I can see why my original solution, not posted here, of 54.9954, was because I was not using a calculator and was doing arithmetic by hand, obviously not too efficient
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 23, 2010

### IronBrain

Bump...100 views and no replies

3. Jan 24, 2010

### IronBrain

WTF BUMP, I am trying to get some help to learn

4. Jan 24, 2010

However frustrated you may feel, swearing at the forum is not going to encourage members to help you.

That said, with problems of this kind you might try plotting the equations as graphs. Do you have access to a spreadsheet program?

This method won't give really accurate answers, but it can help to check your understanding and can provide an approximate check on your mathematical solutions.