Speed Average Velocity Change Question

[Doc Al]

Nope, it's much simpler than that. If "d" is the one way distance, what's the round trip distance?

 

[Kildars]

Nope, it's much simpler than that. If "d" is the one way distance, what's the round trip distance?

2d ;) .

1234567890

 

[Doc Al]

Yes (finally!)

Now use the total time and distance to find the average speed.

 

[Kildars]

Well I did d(1/5.6 + 1/3.4)

I got d(.472).. but I don't know a number for distance so I can't multiply that out..

:(

 

[Doc Al]

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.

 

[Kildars]

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)

 

[Doc Al]

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:
\frac{2a}{a} \ne a

 

[Kildars]

OK.

OK.


Not OK. What happened?

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

(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.

 

[Doc Al]

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!

 

[Kildars]

I didn't mean the division issue, I meant the whole problem.. :)