Homework Help: physics

Posted by Alec on Monday, November 26, 2012 at 11:36pm.

During winter break, the police department in your hometown asks you to consult in the investigation of a traffic accident that occurred in a large parking lot. It seems that a car was backing out of a parking space when a pickup truck plowed straight into it from behind. The driver of the pickup truck claims that he was going 10 mph but police would like you to perform an analysis to verify this. The car was moving 5 mph as it backed up in a straight-line path (three witnesses inside the car independently gave that information so police are trusting it for now). When the two vehicles collided, both drivers immediately slammed on the brakes to lock the wheels. The bumpers crumpled and the two vehicles stuck together. Skid marks show that they traveled 5 meters forward in a straight line before coming to rest. The police have already performed some measurements: the mass of the car is 1500 kg, the mass of the truck is 2500 kg, and the coefficient of kinetic friction between the tire rubber and the asphalt of the parking lot is 0.49

physics - Sierra, Monday, November 26, 2012 at 11:45pm
there's no question, just a bunch of information about the situation
physics - Alec, Monday, November 26, 2012 at 11:58pm
They want you to decide whether or not hte truck driver was actually going 1o mph or not.
What I have so far is v2=(m1v1)/-m2
or v2=(1500kg x 16.093 km/hr)/-2500 kg
which gives -9.6558
However, that seems to simplistic so it's probably super wrong.
physics - Alec, Monday, November 26, 2012 at 11:59pm
They want you to decide whether or not hte truck driver was actually going 1o mph or not.
What I have so far is v2=(m1v1)/-m2
or v2=(1500kg x 16.093 km/hr)/-2500 kg
which gives -9.6558
However, that seems to simplistic so it's probably super wrong.
physics - Elena, Tuesday, November 27, 2012 at 8:53am
m₁=1500 kg, m₂=2500 kg,
v₁=5mph=2.24 m/s, claimed velocity is u=10 mph=4.47 m/s,
real speed of the pickup is v₂=?
μ=0.49, s=5 m.

Law of conservation of linear momentum
m₁•v₁+m₂•v₂=(m₁+m₂)•v,
law of conservation of energy
KE -> W(fr)
(m₁+m₂)•v²/2= μ•(m₁+m₂)•g•s.
v=sqrt(2•μ•g•s)=sqrt(2•0.49•9.8•5)=6.93 m/s.
v₂={(m₁+m₂)•v - m₁•v₁}/m₂=
={(1500+2500)•6.93 -1500•2.24}/2500 =9.74 m/s=24.8 mph
24.8 mph > 10 mph

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