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An automobile and train move together along
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
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 -

    s₁=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.
    t= t₁+t₂+t₃=7.2+63.8+18.2 = 89 s.
    S=v •t=28.8•89=25 63.
    Δs=S-s =2563.2-366.2=2197 m

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