# Help With Problem About 1-D Motion

Okay, so I thought I was on the right track with this problem, but clearly I haven't been getting the right answer!

The problem is as follows :

A physics instructor drives down to St. Louis to see his mom. Coming back, he drives the first half the distance at 65 km/hr and the second half of the distance at 100 km/hr. What is his average speed coming back from St. Louis ?

Okay, so I was trying to determine the t for the two distances and did the following work :

Well I was going to just post a scan of my work, but it says I can't so I'll just give some of what I did...

---

I used t=x/v to solve for the initial t and final t.

So... t initial = x initial / 65km/hr and t final = x final / 100km/hr

Then I plugged those into the equation for average velocity = change in x / change in t

So... V avg = (x - x initial) / [(x / 100) - (x initial / 60)]

---

I think I've gotten to the right point, but if the initial distance and final distance are the same I would just get zero, so I know I'm doing something wrong. Any help would be appriciated! :)

Related Introductory Physics Homework Help News on Phys.org
K, unless I'm very useless at the moment and dumb which happens from time to time... at first, i thought since there's 2 half distances... you add the 2 speeds, and divide them by 2.... giving 82.5km/h but i figured that was wrong.

The other way I tried was...

the total distance is 'x' and V_avg = dist/time

since you dont know time... you have to add in

V_avg = x / [(dist/2)/V1 + (dist/2)/V2]

= x / [(x/2)/65 + (x/2)/100 ]

= 78.78km/h

For some reason that seems wrong, but I gave it a try.

Hello.
In your equation, "V avg = (x - x initial) / [(x / 100) - (x initial / 60)]" you also said that the initial distance x and final distance are the same then. If that was so, then the numerator would be zero. In your case, the initial "x_initial" would be 0 and x would be the total distance traveled.

==========================
By definition of average speed, v=(total distance traveled)/(total time traveled)
Let x = total distance traveled, let x1 = distance of first half, x2=distance of second half, and x1=x2
Let t = total distance traveled, let t1= time of first half traveled, t2= time of second half
Let v1=velocity of first half, v2= velocity of second half

We know that v1=x1/t1, and v2=x2/t2
Then, 65km/hr=(x1/t1) and 100km/hr=(x2/t2). Since x1=x2, then substitute we get:
100km/hr*t2=65km/hr*t1
We can solve for t1=(100/65)*t2 (Eq 1.)

Thus, the average speed would be (x1+x2)/(t1+t2)=(x1+x2)/((100/65)*t2+t2)
by substituting (eq. 1 for t1). Continue from here, we get
t2=x/(2*100km/hr) (Eq. 2)
average speed = x/((100/65)+1)*(x/(200))...
after some simplification, then average speed = 200/(100/65+1)

Last edited:
Dick