# Average speed

1. Feb 5, 2009

### rAz:DD

1. The problem statement, all variables and given/known data
A mobile is moving straight as following: the first portion of the road, equal to 10% of the total distance with the speed v1=10/s , and the rest with the speed v2=30m/s.
The average speed for the entire trip is?
a)25m/s b)22.5m/s c)20m/s d)15m/s

2. Relevant equations

$$\Delta v = \frac{\Delta d}{\Delta t}$$

3. The attempt at a solution

I calculated the times passed for both distances, replaced them in the average speed's formula and got 25 m/s, which is a wrong answer according to the key.

Last edited: Feb 5, 2009
2. Feb 5, 2009

### Staff: Mentor

I think the key is wrong. The correct approach is as you described: calculate the time needed for each portion, add them together, and divide the total distance by the total time.

First part: dist = .1D, rate = 10 m/sec, time = .1D/(10 m/sec) = .01D
Second part: dist = .9D, rate = 30 m/sec, time = .9D/(30 m/sec) = .03D

Average speed = (total distance) / (total time) = D/(.01D + .03D) = D/.04D = 1/.04 = 25 m/sec.

3. Feb 5, 2009

### Staff: Mentor

BTW, the equation you show as relevant is meaningless. dv is differential velocity and d/dt is the differentiation with respect to t operator. The first represents some quantity and the second indicates an operation to perform.

4. Feb 5, 2009

### rAz:DD

Thanks for the answer
And sry for that dv/dt part, i wanted to type the greek letter delta and i didn't know whether tex codes are active or not.

5. Feb 5, 2009

### Staff: Mentor

Yes, they are active. For the upper case delta (looks like a triangle), use \Delta
$$\Delta v = \frac{\Delta d}{\Delta t}$$

6. Feb 5, 2009

### HallsofIvy

Staff Emeritus
Say the road is 100 m long. Then 10% is 10 m and, at 10 m/s, that requires 1 second. The second part is 90 m long and, at 30 m/s, that requires 3 seconds, making a total of 100 m in 4 seconds. That is an average speed of 25 m/s.

What answer does the key give?

7. Feb 5, 2009

### Office_Shredder

Staff Emeritus
The problem is here

It's probably some really exotic unit of distance that was left out.

In other news, I'll second (or fourth at this point) the 25m/s crowd