# Average velocity problem

1. Feb 18, 2008

### bassplayer142

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

A car takes a trip at 20m/s for a certain time interval. At any point in the trip (assuming it doesn't matter), the car takes a 10 minute break with no velocity. The average velocity for the trip is 15m/s The problem asks for the Total time and the distance travelled.

2. Relevant equations

x=vt
v(avg)=1/2(v+v_f)

3. The attempt at a solution

I guess I'm just confused about the rules in using the average velocity equation. This problem should be easy it seems. I have worked it out and I don't have an answer on if what I did is right or not. This is what I did.

T(total)=t(interval)+600(seconds)

x=v*t
x=v*(T-600(seconds))
x=20T-12000(meters)

2*V(avg)=v_o+v_f
=(x_o+x_f)/T=2*X/T

v(avg)=x/T Which makes me think this wasnt even necessisary since it is a general eq.

Plugging in x into the last equation you get

v(avg)*T = 20T-12000(m)
15*T=20T-12000(m)

T=2400s

Therefore x = 36000m

If this is right I would be helpful to know. If not a hint on how to use that stupid average formula would be nice.

2. Feb 18, 2008

### <---

That's the answer I get.
So unless I'm wrong also, you are correct.

3. Feb 19, 2008

### kamerling

The answer is indeed right. The equation v(avg)=1/2(v+v_f) is only valid for constant acceleration. Somehow you used the equation to derive v(avg) = x/T, but this equation
is just the definition of v(avg). The average speed is the total distance travelled divided by the total time taken, and this is always valid. So you could have used this immediately.