- #1
IronBrain
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Homework Statement
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:
You are arguing over a cell phone while trailing an unmarked police car by 37 m. Both your car and the police car are traveling 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:
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.
Homework Equations
Question A:
Code:
(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 traveled 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 traveled 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.
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 traveled by both cop car and my car for this duration. As shown here
Formula to find my distance traveled
[itex]Dist = Speed x Time = 30.55 m/s x 1.9 = 58.045 m[/itex]
Distance traveled by cop car de-accelerating
[itex]D(t) = -2.5t^2+30.55t = -2.5(1.9^2)+30.55(1.9) = 49.02[/itex]
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
[itex]v(t) = -5(1.9)+30.55 = 21.05 [/itex]
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
[itex] -5(t)+21.05[/itex]
Cop's Position Formula
[itex]-2.5(t^2)+21.05(t)+27.975[/itex]
My Car's Velocity Formula
[itex]-5(t) + 30.55[/itex]
My Car's Position Formula
[itex]-2.5(t^2)+30.55(t)[/itex]
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.
[itex]-2.5(t^2)+30.55(t) = -2.5(t^2)+21.05(t)+27.975[/itex]
[itex]30.55(t)-21.05(t) = 27.975[/itex]
[itex]9.5t=27.975[/itex]
[itex]t = 2.9394 s[/itex]
Plugging this into the formula for my velocity
[itex]-5(2.9394)+30.55 = 15.853 m/s[/itex]
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