Speed Average Velocity Change Question

In summary, the person walks first at a constant speed of 3.40 m/s along a straight line from Point A to Point B and then back along the line from B to A at a constant speed of 5.60 m/s.
  • #36
Doc Al said:
You don't need a number for the distance (or the time for that matter). Take the expression for total distance that you found and divide it by the expression for total time.

Ok

Expression for total distance = 2d
Expression for total time = d(.472)

2d / d(.472)

d / (.472)
 
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  • #37
Kildars said:
Expression for total distance = 2d
Expression for total time = d(.472)
OK.
2d / d(.472)
OK.

d / (.472)
Not OK. What happened?

Ahh... I see what you might have done. Note, in general:
[tex]\frac{2a}{a} \ne a[/tex]
 
  • #38
Doc Al said:
OK.

OK.


Not OK. What happened?

Ahh... I see what you might have done. Note, in general:
[tex]\frac{2a}{a} \ne a[/tex]

(2)(.472) = 4.23

Correct answer, wow thanks.

Could you explain a little bit more how that works? Because without your assistance I wouldn't have been able to get that correct.
 
  • #39
Not sure what I could explain about it further. It looked to me that you were thinking of subtraction instead of division: 2a - a = a, but 2a/a = 2. Big difference.

More important is to revise your problem solving strategy. As this problem illustrates, sometimes you don't need information that you think you need. When in doubt, label the unknown and keep on going!
 
  • #40
I didn't mean the division issue, I meant the whole problem.. :)
 

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