Please help me answer this question.?

An automobile starting from rest speeds up to 40ft/sec with a constant acceleration of 4ft/sec^2 runs this speed for a time, and finally comes to rest with a deceleration of 5ft/sec^2. If the total distance traveled is 1000ft find the total time required ?

2 Answers

Relevance
  • Anonymous
    4 weeks ago
    Favorite Answer

    d1 = V^2/2a1 = 40^2/8 = 200 ft

    t1 = V/a1 = 40/4 = 10 sec

    d2 ) V^2/2a2 = 40^2/10 = 160 ft

    t2 = V/a2 = 40/5 = 8 sec

    d3 = D-d1-d2 = 1000-(200+160) = 640 ft

    t3 = d3/V = 640/40 = 16 sec

    total time t = t1+t2+t3 = 18+16 = 34 sec

  • 4 weeks ago

    it takes t= v/a = 40/4 = 10 s to accelerate and t = 40/5 = 8 s to come to a stop.  The distance moved is v/2 * 2 = 20 * 18 = 360 ft

    there are now 640 feet left to be moved at 40 ft/s which takes t = s/v = 640/40  = 16 s.  The total time = 18+16 = 34 s

    • ...Show all comments
    • Andrew Smith
      Lv 7
      4 weeks agoReport

      Dianna, no problem.  You can only award one answer.  But given that both answers are correct the only difference is how they are laid out.  Since they have gone to only showing the first FOUR lines of an answer it has seriously degraded my ability to give a good well laid out answer.

    • Login to reply the answers
Still have questions? Get your answers by asking now.