# I am confused on how to derive time, and velocity.

1. Feb 3, 2016

### Eclair_de_XII

1. The problem statement, all variables and given/known data
"An automobile travels on a straight road for 40-km for 30 km/h. It then continues in the same direction for another 40-km at 60 km/h. (a) What is the average velocity of the car during the full 80 km trip? (Assume that it moves in the positive x direction.) (b) What is the average speed?"

$d_1=40 km$
$d_2=80 km$
$v_1=30 km/h$
$v_2=60 km/h$

2. Relevant equations
$v=v_0+at$
$x-x_0=v_0t+\frac{1}{2}at^2$
$v^2 = v^2_0+2a(x-x_0)$
$x-x_0=\frac{1}{2}(v_0+v)t$
$x-x_0=vt-\frac{1}{2}at^2$

3. The attempt at a solution
Basically, I am trying to solve for time, but I am getting conflicting answers. This is the method I am typically accustomed to:

$(\frac{1 h}{30 km})(40 km) = \frac{4}{3}h$

This is an entirely different result from what I get from equation four on the list.

$(40 km) - (0 km) = \frac{1}{2}(0+30km/h)t$
$80km=(t)30km/h$
$t=\frac{8}{3}h$

As a result, I cannot solve the two parts for this problem, which this book tells me, is 40 km/h for each one. Additionally, I feel the need for a confession... This was the same book I had earlier complained about not having the proper formulae. It was not that it didn't have them; it's just that I was looking in the wrong places. My professor skipped around the chapters for her curriculum. After I had dropped the class, I tried to follow that curriculum, but the formulae I needed were in the chapters she had skipped when I was still taking her class. Anyway, I don't understand why I'm getting such conflicting answers when I try to solve for time.

2. Feb 3, 2016

### SteamKing

Staff Emeritus
Most of your relevant equations are not suitable for this type of problem. Some of them are for accelerated motion, which is clearly not the case here.

This problem can be solved by applying the plain old equation: distance = rate * time

If you travel 20 km at 40 km/hr, it's going to take you 20 km / 40 km/hr = 1/2 hr to go that distance.

Re-read the two parts to this problem and try again. It's a lot simpler than it looks.

3. Feb 3, 2016

### SammyS

Staff Emeritus
Those equations, including the fourth one, are for constant acceleration.

Physics is about much more than simply picking out equations which might happen to have the relevant quantities in them.

4. Feb 5, 2016

### Eclair_de_XII

I would figure otherwise, that I would average the two speeds together to get 45 mph; but the book states that it's 40 mph. It just doesn't make any sense.

Yeah, no kidding. You actually have to think about which equations to use, how to use them, and whether you use them at all. It's not simply just plugging in numbers into the appropriate equations, like chemistry was for me.

5. Feb 5, 2016

### Eclair_de_XII

Wait, I think I just figured it out!

$\frac{1hr}{30km}(40km) = \frac{4}{3}hr$
$\frac{1hr}{60km}(40km) = \frac{2}{3}hr$
$\frac{4}{3}hr+\frac{2}{3}hr=2hr$
$(40km)+(40km)=80km$
$\frac{80km}{2hr}=40\frac{km}{hr}$

I'll try working on those online problems someone took the liberty of linking me to; that is, if I'm able to find some simple problems. The ones on the front page are much too complicated for me, I think.