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July 23, 2014

July 23, 2014

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

- physics -
**Sierra**, Monday, November 26, 2012 at 11:45pmthere's no question, just a bunch of information about the situation

- physics -
**Alec**, Monday, November 26, 2012 at 11:58pmThey 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 11:59pmThey 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 8:53amm₁=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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