Friday

November 28, 2014

November 28, 2014

Posted by **m** on Tuesday, May 21, 2013 at 8:13am.

parallel paths at 28.8 m/s.

The automobile then undergoes a uniform

acceleration of −4 m/s

2 because of a red light

and comes to rest. It remains at rest for 63.8 s,

then accelerates back to a speed of 28.8 m/s

at a rate of 1.58 m/s

2

.

How far behind the train is the automobile

when it reaches the speed of 28.8 m/s, assuming that the train speed has remained at

28.8 m/s?

Answer in units of m

- physics -
**Elena**, Tuesday, May 21, 2013 at 11:30ams₁=v²/2a₁=28.8²/2•4=103.7 m.

s₁=a₁t₁²/2 =>

t₁=sqrt(2s₁/a₁) =sqrt(2•103.7/4) =7.2 s.

t₂=63.8 s.

s₃=v²/2a₃=28.8²/2•1.58=262.5 m

s₃=a₃t₃²/2 =>

t₃ =sqrt(2s₃/a₃) =sqrt(2•262.5/1.58) =18.2 s.

Automobile:

s=s₁+s₃=103.7+262.5=366.2

t= t₁+t₂+t₃=7.2+63.8+18.2 = 89 s.

Train:

S=v •t=28.8•89=25 63.

Δs=S-s =2563.2-366.2=2197 m

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