A person takes a trip: Find the time spent on the trip, etc.

  • #1

Homework Statement


A person takes a trip, driving with a constant of 89.5 Km/h, except for a 22.0-min rest stop.
If the person's average speed is 77.8 Km/h,
(a) how much time is spent on the trip and
(b) how far does the person travel?

This is problem 13, Chapter 2 from Serway Physics for scientists and engineers Volume 1 9th edition

Homework Equations




The Attempt at a Solution


Serway Physics 2_13001.jpg
Serway Physics 2_13002.jpg

[/B]
Is my solution correct?
 

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Answers and Replies

  • #2
verty
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When you wrote $$V_{avg} = {89.5 t_{a \to b} + 89.5 t_{b \to c} \over t_{a \to b} + t_{b \to c} + t_{rest}}$$, that was correct but I think they mean to include the rest stop in the time taken.
 
  • #3
kuruman
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I disagree with your numbers. It makes no sense to to split the time interval into 3 parts because it doesn't matter when the driver rests. For all you know, the driver can start 22.0 min late and have no rest stop and it wouldn't make a difference to the average speed and the distance traveled. The average speed is $$v_{avg}=\frac{D}{t_{rest}+t_{drive}}$$where ##D## is the total distance traveled. You also have a second equation relating ##D## and ##t_{drive}##. The two equations comprise a system of two equations and two unknowns, ##D## and ##t_{drive}##, that you have to solve.
 
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  • #4
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Yes, your solution is correct with the small correction that 77.8 * (22/60) = 28.53 not 28.008. That’s just some sort of round off error, but I can’t spot where or why it is so big.

I think you could do this much more succinctly effectively jumping right to the middle of your calculation by appropriate logic. What you arrive at third line from the bottom of part a:

(tdrive+ 22 min) * vavg = tdrive * vdrive

seems readily apparent and could easily have been written as the starting point of the solution.
 
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  • #6
kuruman
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Your numbers agree with mine.
 
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