# How to solve for distance give average velocity of magnitude

1. Sep 7, 2015

### bthebiologist

I have changed the values, and the homework also changes every time I submit a response, I would just like to see an example worked out, so change the values as well if you want!

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

A person takes a trip heading south with an average velocity of magnitude 120.0 km/h except for a 0.50h stop. Overall average velocity for the trip is 60.0 km/h. How far in km is the trip?

2. Relevant equations
Part of the homework was to come up with the relation between the 3 times, which I got right at
Toverall=-Tbreak+Ttrip

However I do not understand how this would help solve the problem.

3. The attempt at a solution
I have tried many different attempts, but the one I feel should work, just not sure how would be

120.0km/h = (60.0km/x(h))+(0.0km/(0.6h))

however this was not right, also tried using t=d/r

2. Sep 7, 2015

### Borg

Welcome to PF.
You have two things that are equal here. One involves the actual speed of the train with the stop. The other involves the average speed.

3. Sep 7, 2015

### BvU

Hello Btb, welcome to PF !

Physicists (and mathematicians) have another way of ordering things. They say:
How do I call things?
What do I know ?
What equations do I have ?
What equations do I have to solve ?

Well, you have $$T_{\rm over all}, \quad T_{\rm break}, \quad T_{\rm moving}, \quad v_{\rm over all}, \quad v_{\rm break}, \quad v_{\rm moving}$$
What you know $$v_{\rm over all} = 120 \, {\rm mph}, \quad T_{\rm break} = 0.5\, {\rm h}, \quad v_{\rm break} = 0 \quad v_{\rm moving} = 120 \, {\rm mph}$$
 sorry, km/h instead of mph. Makes the car possibility even more reasonable
Equations you have $$T_{\rm over all} = T_{\rm break} + T_{\rm moving}$$ $${\rm Distance} = v_{\rm over all}\, T_{\rm over all}$$
With 7 variables, of which 4 known and 2 equations, you're still one short. What can it be ? Aha ! $${\rm Distance} = v_{\rm moving}\, T_{\rm moving } + v_{\rm break} \, T_{\rm break}$$
So all we have to solve for -- after substitution -- is one of the unknown T. From: $$v_{\rm moving}\, T_{\rm moving} = v_{\rm over all}\, T_{\rm over all} \quad {\rm and }$$ $$T_{\rm over all} = T_{\rm break} + T_{\rm moving}$$
I'm sure you can work that out.
Borg assumes you travel by train. So no speeding tickets.

Last edited: Sep 7, 2015
4. Sep 7, 2015

### Borg

Nice catch. Clearly I didn't read close enough.

5. Sep 7, 2015

### bthebiologist

Okay so I think, but may be wrong, that Voverall should be 60.0km/h not 120km/h, so my equations would be

(60.0km/h)(xh-0.5h)=(120km/h)(xh)
and
xh=(xh-0.5)+0.5h

I am guessing that the substitution would be to put (xh-0.5h)+0.5h into the equation where xh is.
Therefore

(60.0km/h)[{(xh-0.5h)+0.5h}-0.5h)=(120km/h)[(xh-0.5h)+0.5h]

Which would reduce to (60.0km/h)(-0.5xh^2+0.25h^2-0.25h^2)=(120km/h)(0.5xh^2-0.25h^2)

Then (60km/h)(-0.5xh^2)=(120km/h)(0.5xh^2-0.25h^2)

but x's would cancel out leaving me with 60km/h=120km/h(+0.25h^2)?

I guess I am just lost as to what we are subsituting, the whole question makes sense to me we are just trying to figure out how long we would need to travel at 120km/h to make our average speed 60km/h since grandma had to use the bathroom for half an hour

So to me this make sense to me -> 60km/h(t)=(120km/h)(t-0.5)+(0km/h)(0.5)
which is 60km/h(t)=(120km/h)(t-0.5)
but all the hours and kilometers are confusing me. is 60km/h(t) = 60km(th) and (120km/h)(t-0.5)= 120km(th)-60km?

6. Sep 7, 2015

### Ray Vickson

You are making it WAAAY too complicated and lengthy. Just let x = distance travelled (in km) and t = travelling time (in hours); this is the time actually spent moving. The total trip time is T = t + 0.5. We are given 120 = x/t and 60 = x/(t+0.5), so we have two equations in the two unknowns x and t. We can use the first equation to write x in terms of t: x = 120 * t, and putting that into the second equation gives an equation involving t alone.

7. Sep 7, 2015

### bthebiologist

Okay so we have 60=(120*t)/(t+0.5)
60 (t+0.5)=120*t
60t+30=120t
30=60?
Not sure how to keep t since it would naturally cancel.

8. Sep 7, 2015

### Borg

There is no 't' with the 30 so the t's don't all cancel.

9. Sep 7, 2015

### bthebiologist

Ah crap. So 30=60t so 0.5=t

So 120=x/0.5 -> 60= x
And 60=x/(0.5+0.5) -> 60=x

So is my total time = 1hr, drive time =0.5h
With distance being 0.5h (120km/h) = 60km traveled?

10. Sep 7, 2015

### Borg

Yes, that's correct.

11. Sep 7, 2015

### bthebiologist

OH GOD THANK YOU JESUS! Thank you guys (gals?) So much! Onto the next question... :/

12. Sep 7, 2015

### BvU

One more reason to work with symbols as long as reasonable. There were two equations left over with two unknowns, remember?$$v_{\rm moving}\, T_{\rm moving} = v_{\rm over all}\, T_{\rm over all} \quad {\rm and }$$ $$T_{\rm over all} = T_{\rm break} + T_{\rm moving}$$ let's get rid of $T_{\rm over all}$ and write $$v_{\rm moving}\, T_{\rm moving} = v_{\rm over all}\, \left ( T_{\rm break} + T_{\rm moving} \right )$$ so that
$$\left ( v_{\rm moving} - v_{\rm over all} \right ) \, T_{\rm moving} = v_{\rm over all} \, T_{\rm break}$$
and therefore $$T_{\rm moving} = T_{\rm break} \, { v_{\rm over all}\,\over v_{\rm moving} - v_{\rm over all} }$$
And now that you have an unknown on the left and all the stuff on the right is known, it's time to fill in some stuff.

Ray already said it: don't mix up values and dimensions.

-- slow typst. I see you're all done already but some example in dealing with equations may be useful, nevertheless.