Average speeds of back and forth trip

  • Thread starter Thread starter 1MileCrash
  • Start date Start date
  • Tags Tags
    Average
AI Thread Summary
The discussion revolves around calculating average speeds for a trip from San Antonio to Houston and back, with varying speeds. The average speed from San Antonio to Houston is calculated as approximately 68.28 km/h, while the return trip averages 72.5 km/h. There is confusion regarding the interpretation of "half the time" versus "half the distance," leading to a debate about the correct application of the average speed formula. Clarifications are made that average speed is determined by total distance divided by total time, not simply by averaging the speeds. The conversation concludes with a better understanding of the relationship between distance, speed, and time in these calculations.
1MileCrash
Messages
1,338
Reaction score
41

Homework Statement



You drive on Interstate 10 from San Antonio to Houston. Half the time at 55 km/h, and the other half at 90 km/h. On the way back, you travel half the distance at 55 km/h and the other half at 90km/h.

What is the average speed of the trip from San Antonio to Houston?
What is the average speed of the return trip from Houston to San Antonio?
What is the average speed of the whole trip?


The Attempt at a Solution



From San Antonio to Houston, half the time is traveled between the two speeds.

So:

d / (1/2)(d/55) + (1/2)(d/90)
68.27586 km/h average speed.

From Houston to San Antonio, half the distance is traveled at each speed.

((1/2)55t + (1/2)90t) / t
=(1/2)55 + (1/2)90
=
72.5 average speed.


My book seems to have the two answers reversed. Did I do these correctly?
 
Physics news on Phys.org
1MileCrash said:

Homework Statement



You drive on Interstate 10 from San Antonio to Houston. Half the time at 55 km/h, and the other half at 90 km/h. On the way back, you travel half the distance at 55 km/h and the other half at 90km/h.

What is the average speed of the trip from San Antonio to Houston?
What is the average speed of the return trip from Houston to San Antonio?
What is the average speed of the whole trip?

The Attempt at a Solution



From San Antonio to Houston, half the time is traveled between the two speeds.

So:

d / (1/2)(d/55) + (1/2)(d/90)
68.27586 km/h average speed.

From Houston to San Antonio, half the distance is traveled at each speed.

((1/2)55t + (1/2)90t) / t
=(1/2)55 + (1/2)90
=
72.5 average speed.My book seems to have the two answers reversed. Did I do these correctly?

You said, "From San Antonio to Houston, half the time is traveled between the two speeds."

I added the extra -- necessary -- parentheses in the following.
d / ((1/2)(d/55) + (1/2)(d/90))
What is (d/2)/55 ? ... It's the time required to travel a distance d/2 at a speed of 55...
etc ...

d/(total time) is average speed.

So this is not the case of: "half the time is traveled between the two speeds".
 
I am not sure what you are saying..

Does my work not agree with total distance / total time? Why are you telling me that?

Also, why isn't it a case of traveling half the time at each speed? That's explicitly what the problem says.
 
1MileCrash said:
...

Also, why isn't it a case of traveling half the time at each speed? That's explicitly what the problem says.

Because d/2 is half the distance !

So you are finding the time for half the distance at 55 + half the distance at 90.
 
Now I'm really confused. If time is (d/r), (1/2)(d/r) or (d/2r) is half of the time is it not??

Wait, I think I'm starting to get it. d/r is the time spent traveling d at r, so half of it is not half of the total time, only half of the time spent traveling d at r.

Thanks!
 
Last edited:
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top