Avg. speed, distance, and time

  • Thread starter dayzie
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  • #1
dayzie
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Homework Statement



A car undergoes a displacement of 1200m @ 90degrees, follow by another displacement of 900m @ 0degress.

Calculate the average speed for the whole trip and the time to return to the starting point in the shortest possible distance at 20.0m/s


Homework Equations


Avg speed would be total displacement/total time
2100m/210s=10.0 m/s, which is not right, it should be 12.7m/s and I don't know how to arrive at that figure?

The time for the second equation should be 75.0s. How do I get there?
If the total time that I have for the whole trip is 210s? Is that even the correct time?


The Attempt at a Solution

 

Answers and Replies

  • #2
JesseC
251
2
Draw yourself a picture, maybe the second part will become clearer.

For the first part, have you got the right total time?
 
  • #3
JaredJames
2,817
22
There's some details missing from your question. Where is the total time shown?

For the second part, if you draw a diagram you will see it is a right angle triangle. You need to use pythagoras theorem to work out the shortest route back the start (a2+b2=c2).

Once you have that distance, you use the speed of 20m/s given in the question to calculate the time to travel back.
 
  • #4
dayzie
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The time is not shown, we have to figure that out.
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
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Homework Statement



A car undergoes a displacement of 1200m @ 90degrees, follow by another displacement of 900m @ 0degress.

Calculate the average speed for the whole trip and the time to return to the starting point in the shortest possible distance at 20.0m/s


Homework Equations


Avg speed would be total displacement/total time
2100m/210s=10.0 m/s, which is not right, it should be 12.7m/s and I don't know how to arrive at that figure?
The dispacement is not 2100m. The car has moved along two legs of a right triangle. Its displacement if the straight line distance from start point to end point. Use the Pythagorean theorem to find the length of the hypotenuse of that right triangle.

How did you get "210 s"? I see nothing said here about the speed on these two legs nor about the time. Are we to assume the speed was also 20 m/s? Please post the entire question.

The time for the second equation should be 75.0s. How do I get there?
If the total time that I have for the whole trip is 210s? Is that even the correct time?
Where did you get those numbers? They are not given in the problem.


The Attempt at a Solution

 

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