physics--drwls

A record of travel along a straight path is as follows:
1. Start from rest with constant acceleration of 2.77 m/s^2 for 15 s
2. Constant velocity for the next 2.05 min
3. Constant negative acceleration -9.47 m/s^2 for 4.39 s.

(a) What was the total displacement for complete trip?
(b) What were the average speeds for legs 1,2,and 3 of the trip as well as for the complete trip.

Answers in the book: (a) 5510 m (b) 20.8 m/s, 41.6 m/s, 20.8 m/s, 38.7 m/s

physics - drwls, Friday, September 7, 2007 at 1:09am
Just do it one step at a time and add the distances traveled. During part 1, the distance travelled is
(1/2) a t^2 = (1/2)(2.77)(225) = 311.6 m
The final speed attained is (2.77)(15) = 41.55 m/s. The average speed duering that interval is half of that, or 20.8 m/s
During part 2, the distance traveled is
41.55 m/s*2.05 min*60 sec/min = 5110 m/s
During part 3, the speed drops from 41.55 m/s to
41.55 - (9.47)(4.39) = 0
The average speed during this interval is 41.55/2 - 20.78 m/s and the distance travelled is 20.78*4.39 = 91.2 m/s
Total distance travelled = 312 + 5110 + 91 = 5513 m. In your book's answer, they rounded off some of the numbers differently. You can only trust the first three significant figures, hence the 5510 m answer.

How would I get the average speed for the complete trip? (38.7 m/s)

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  1. average speed entire trip?

    average speed= totaldistance/totaltime

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  2. Thanks. I thought it was more complex than that.

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  3. joe

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