Average velocity ?

1. Nov 8, 2004

pseudonewtonian

Average velocity ??????

You're driving an interstate 10 from Sn Antonio to Houston half the time at 55km/hr and the other half at 90km/hr . What is your average speed a) from San Antonio to Houston b) from Houston back to San Antonio c) for the entire trip ? What is your average velocity for the entire trip ? Indicate how average velocity can be found on a sketch of x versus t ?

HI i'm new to the forums and this question has me all confused..... Any help is greatly appreciated. Thanks in advance.

2. Nov 8, 2004

Galileo

The average speed over some time interval T is (distance travelled)/T
You know you travelled 55 km/hr for a period of T/2
and 90 km/h for the remaining period of T/2.
With this you can find the total distance travelled.

3. Nov 8, 2004

HallsofIvy

Staff Emeritus
If you do not know the answers to a) and b) you have a serious problem! Run, do not walk to your instructor and throw yourself on his/her mercy.

The answer to c) depends upon whether you really mean "velocity" rather than "speed" (do you know the difference?). If it really is "velocity" (and you know the distinction between "speed" and "velocity") the problem is trivial.

If you really mean the "average speed", then you will have to know the distance between San Antonio and Houston. That CANNOT be calculated from the information given here- you may have to consult a map! I notice that one speed is given in "miles per hour" and the other in "kilometers per hour". If you find the distance in miles, you will want to convert " 60 kilometers per hour" to "__ miles per hour". If you find the distance in kilometers, you will want to convert the other way. Once you know the distance, how long (time) does it take to drive from San Antonio to Houston at 55 mi/hr? How long does it take to drive from Houston to San Antonio (same distance of course) at 60 km/hr. What is the total time taken to drive the total distance (twice the distance from San Antonio to Houston)?

4. Nov 8, 2004

UrbanXrisis

:rofl: :tongue:

5. Nov 9, 2004

pseudonewtonian

consult a map... ? Thats not given in the problem.. and i got a serious problem u say ? Well u might be right. I still do not understand what to do... This sum is from Fundamentals Of Physics - Resnick Halliday Walker 1st chapter.. and by the way i missed out on some info..... ie.. On the way back from Houston u drive at 55km/hr for half the distance and 90km/hr for the other half.... this should've been there before the questions

6. Nov 9, 2004

ek

A and B really are pretty darn easy questions. They're not even really physics.

You have six apples and I have four apples, what is the average number of apples we each have?

7. Nov 9, 2004

pseudonewtonian

ya ya i get it ek... and i know the difference between physics sums and apples.... thats ok... i know how to do it that way but is that quite right physically ????? Thats my problem I dont get the physical significance of it otherwise average is kid's play.. btw this is not my homework or anything so i aint being lazy.... kinematics is driving me mad

8. Nov 9, 2004

ek

There's nothing more to it than that.

55 + 90 / 2

So if the distance between the two cities was 290km, then it would take 4 hours.

55 km/hr for 2 hrs = 110 km
+
90km/hr for 2 hrs = 180 km
290 km in four hours.
290/4 is 72.5km/hr, just like the original equation.

9. Nov 9, 2004

primarygun

May I know how old is a Grade K-12 student?

10. Nov 9, 2004

ek

K -> Kindergarten -> 5 years old.

12 -> Grade 12 -> 16/17/18 years old.

K-12 is every age in between.

11. Nov 12, 2004

primarygun

Oh so I am not in G-12 lol

12. Nov 12, 2004

BobG

Make sure whether the question is saying you drove half the time at a given speed or half the distance at a given speed. Half the time is pretty easy - you just take the average of your speeds. For half the distance, you need to find out how long it took you for the first leg, then how long it took you for the second leg.

And, actually, if you play with the equations, you can find the average speed without knowing the distance, even in the second case.

$$t_{tot}=\frac{d}{s_1}+\frac{d}{s_2}$$

or

$$t_{tot}=\frac{ds_2 +ds_1}{s_1s_2}$$

$$s_{ave}=\frac{2d}{t_{tot}}$$

The distance winds up canceling out.