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bby__hae!
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[SOLVED] [attempt?] car accident problem.
http://hypertextbook.com/physics/mechanics/momentum-two-three/accident-reconstruction-1.pdf [Broken]
Pix=Pfx
Piy=Pfy
v^2 = v0^2 + 2 ax
Fk=MkN
W=mg
F=ma
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.
Homework Statement
http://hypertextbook.com/physics/mechanics/momentum-two-three/accident-reconstruction-1.pdf [Broken]
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.
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