# Straight line, fractional difference

1. Sep 15, 2011

### CarEnthusiast

Straight line, "fractional difference"

1. The problem statement, all variables and given/known data

To set a speed record in measured (straight) distance d, a race car must be driven first in one direction (in time t1) and then in the opposite direction (in time t2). (a) To eliminate the effects of the wind and obtain the car's speed vc in a windless situation, should we find the average of d/t1 and d/t2 (method 1) or should we divide d by the average of t1 and t2? (b) What is the fractional difference in the two methods when a steady wind blows along the car's route and the ratio of the wind speed vw to the car's speed vc is 0.0240?

2. Relevant equations

t1 = with wind
t2 = against wind
v1 = d/t1
v2 = d/t2
v1 = vc+0.0240vc
v2 = vc-0.0240vc

"Method 1": vc = (v1+v2)/2
"Method 2": vc = d/(1/2(t1+t2))

3. The attempt at a solution

I understand the principle behind the problem (the physics concept is easy), but I cannot understand how one method is better than the other if they end up equaling the same thing. Also, where does the "fractional difference" come in, and how do you calculate it?

2. Sep 15, 2011

### Modulated

Re: Straight line, "fractional difference"

Do they really "end up equalling the same thing"? You certainly haven't shown that they do - perhaps you should try to manipulate your answers to "Method 1" and "Method 2" to check whether or not this is true.

The "fractional difference" between two numbers is the difference between them divided by the "reference" number - which in this case would be whichever of the two methods correctly predicts the windless speed. (I.e.,

$$\frac{\text{other method} - \text{windless speed}}{\text{windless speed}}$$

)