# 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