[attempt?] car accident problem.

[bby__hae!]

[SOLVED] [attempt?] car accident problem.

Homework Statement

http://hypertextbook.com/physics/mechanics/momentum-two-three/accident-reconstruction-1.pdf

Homework Equations

Pix=Pfx
Piy=Pfy
v^2 = v0^2 + 2 ax
Fk=MkN
W=mg
F=ma

The Attempt at a Solution

01.
Fk = MkN
100 N = Mk (130 N)
0.769 = Mk

car tires = 0.769 --- but then, I'm not sure of the units.


0.769 (0.7) = 0.538

truck tires = 0.538 --- again with the units.

02.
W = 13600 N
W = mg
13600 = m (9.8 m/s^2)
1387.755 kg = m

Fk = MkN
Fk = 0.769 (13600 N)
Fk = 10458.4

F = ma
10458.4 = 1387.755 a
7.536 m/s^2 = a

v^2 = v0^2 + 2 ax
0 = v0x^2 + 2 (-7.536) (8.2)
123.59 = v0x^2
11.117 m/s = speed of car


W = 69700 N
W = mg
W = 7112.245

Fk = MkN
Fk= 0.538 (69.700)
Fk = 37498.6

F = ma
37498.6 = 7112.245 a
a = -5.272 m/s^2

v^2 = v0^2 + 2 ax
0 = v0^2 + 2 (-5.272) (11)
115.984 m/s = v0^2
10.769 m/s = speed of truck

03.
Pfx=Pix --- i know i wrote the other way earlier, but I'm copying my weird notes.
(1387.755) (11.117) cos (33) + (7112.245) (10.769) cos (7) = 7112.245 v
88958.823 = 7112.245 v
v of truck = 12.508 m/s


(1387.755) (11.117) sin (33) + (7112.245) (10.769) sin (7) = 1387.755 v
17736.701 = 1387.755 v
v of car = 12.781 m/s

I asked my teacher and he said he vaguely remembering that the velocities were supposed to be 11?, but he and I couldn't find fault with the set up. I tried recalculating, and I came up with the same answer, which frustrates me a bit.

04.
As for number four, I read the facts given repeated times, but I'm still at a loss for how to approach it. I realize that I'm supposed to use parts 01.-03. though.



Thanks. :D
Sorry, it's a bit long.

 

[Jose Vallin]

The measurement of the distance between the traffic light and the collision point (d = 13.0 m) and the maximum acceleration of the car (a = 3.0 m/s^2) are not consistent with the claim of the car driver. Assuming constant acceleration, which is a pretty good assumption since the mass of the car does not change significantly as a result of the collision, we get:vf = Sqrt(2ad) = Sqrt[ 2(3.0m/s^2)(13.0 m) ] = 8.83 m/s.The percent difference between this speed and the speed of the car just before the collision calculated in question 3 is significant.% Difference = [ 12.781 m/s - 8.83 m/s ]/[ 12.781 m/s + 8.83 m/s ] x 200 = 37%.