# 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

- Anonymous1 month agoFavorite 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

- Andrew SmithLv 71 month 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
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.

- Log in to reply to the answers

Thank you very much