Average Speed With Ambiguous Distance/Time

[onemic]

Homework Statement

A locomotive travels on a straight track at a constant speed of 40 mi/h, then reverses direction and returns to its starting point, traveling at a constant speed of 60 mi/h. What is the average speed for the round-trip?

Homework Equations

_avgS = distance/time
t = d/40

The Attempt at a Solution

_avgS = 2d/(t+40/60(t))
= 2d/(t+(2/3)t)
= 2(t/40)/(t+(2/3)t)
= (t/20)/(t+(2/3)t)
I don't really know what to do after this point. The solution manual has the steps for the solution as:

_avgS = 2d/(t+(2/3)t)
= 80t/(t+(2/3)t) = 48 mi/h

I have no idea how they got to the last step. Any help is greatly appreciated.

 

[maxf11]

In case you don't know how they got the first step, they took the total distance over the total time: it's t+2/3t because he's traveling 1.5 times as fast on the way back. as t represents the amount of time it takes the train to get to the turning point, we can use the formula d = 40t (because distance = speed*time) and substitute into the formula given in the answers: 2d = 80t.

 

[onemic]

Ah, thanks. How did they get the final answer of 48 mi/h?

 

[maxf11]

80t/(5/3t)=240t/5t=48

 

[onemic]

Thank you!

 

[Ray Vickson]

Homework Statement

A locomotive travels on a straight track at a constant speed of 40 mi/h, then reverses direction and returns to its starting point, traveling at a constant speed of 60 mi/h. What is the average speed for the round-trip?

Homework Equations

_avgS = distance/time
t = d/40

The Attempt at a Solution

_avgS = 2d/(t+40/60(t))
= 2d/(t+(2/3)t)
= 2(t/40)/(t+(2/3)t)
= (t/20)/(t+(2/3)t)
I don't really know what to do after this point. The solution manual has the steps for the solution as:

_avgS = 2d/(t+(2/3)t)
= 80t/(t+(2/3)t) = 48 mi/h

I have no idea how they got to the last step. Any help is greatly appreciated.

I much prefer to do it from first principles, in detail and without skipping steps; here is how:

If distance out = distance in = d (in miles), time out is T_o = d/40 (in hours), and time in is T_i = d/60. Total time = T_o+T_i = d(1/40 + 1/60) = (100/2400) d. Round-trip distance is 2d, so average speed is S_avg = 2d/(total time) = 2400/100 = 24 (in miles/hr).

 

[Mark44]

I much prefer to do it from first principles, in detail and without skipping steps; here is how:

If distance out = distance in = d (in miles), time out is T_o = d/40 (in hours), and time in is T_i = d/60. Total time = T_o+T_i = d(1/40 + 1/60) = (100/2400) d. Round-trip distance is 2d, so average speed is S_avg = 2d/(total time) = 2400/100 = 24 (in miles/hr).

You have an error here, Ray, from a lost or misplaced factor of 2. The average speed for the entire trip is 48 mi/hr.

 

[Ray Vickson]

You have an error here, Ray, from a lost or misplaced factor of 2. The average speed for the entire trip is 48 mi/hr.

Thanks. That was an obvious typo that I did not see before pressing the enter key.