# Two vehicles moving in opposite directions bonus problem

1. Jun 23, 2014

### antonisz

1. The problem statement, all variables and given/known data
Vehicle A and Vehicle B are moving in opposite directions on the NJTP. Vehicle A is heading south toward atlantic city while vehicle B is heading north towards Hoboken. In situation 1 and 2 described below, at t = 0s both vehicles are at a distance of separation of 400m and are moving towards each other.

Situation 1: When vehicle A is moving at a constant velocity of 30 m/s and travels a distance of 120 m, the vehicles pass each other on the turnpike.

Situation 2: When vehicle A is moving at a constant velocity of 80 m/s and after a time of 3s, the vehicles pass each other on the turnpike.

For vehicle A label variables as: Via, Vfa, aa, da,ta
For vehicle B label variables as: Vib, Vfb, ab,db, tb

In the ORDER INDICATED:

(a) Calculate the initial velocity of vehicle B or Vib; and

(b) Then calculate the acceleration of vehicle B or ab

Show all work in DETAIL and keep all numerical values to the nearest hundredth value. Draw and label all pictures

Note: You will receive NO CREDIT if you first calculate part (b) and then part (a)

For situations 1 and 2

vib = constant ; vfb ≠ a constant value since its value changes w/ time

ab = constant

2. Relevant equations

3. The attempt at a solution

Situation 1:
Da + Db = 400

Via = 30 m/s
Vfa = 30 m/s
a = 0 m/s2
da = 120 m
ta = ?

Vave = d/ t = vf + vi /2
Vave = da/ta
ta = 4s

db = 280
ta = 4s

Situation 2:
Via = 80 m/s
aa = 0 m/s2
vfa = 80 m/s
ta = 3s = tb
da = ?

Using same equation as in situation 2,

da = 240

db=160
tb = 3s

Situation 1
db = vibtb + 1/2abtb2
280 = vib(4) + 1/2 (ab)(4)2

Situation 2
db = vibtb + 1/2abtb2
280 = vib(3) + 1/2 (ab)(3)2

280 = vi (4) + 8ab
280 - 4bi = 8ab
280 - 4vi/8 = ab

Plug this into the second equation...

160 = vib(3)+ 1/2 (280 - 4vi/8 )9
I solve and get vib = - 4/3 m/s which doesn't make sense...

So i plug that answer back into the two equations above and get ab = 35.67 m/s2 for situation 1 and for situation 2 - 34.67 m/s2.

Thank you for the help
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 23, 2014

### BiGyElLoWhAt

Are you sure you didn't leave anything out? All you know for sure about vib is that it's in the opposite direction of a and changes with time. If I say vib = 1m/s, then you can solve for the acceleration using the time values, but if I say vib =2m/s then you can do the exact same thing, but you'll get a different answer.

Reread the question and make sure you didn't leave out anything important please.

3. Jun 23, 2014

### antonisz

I copy and pasted the file that the professor gave us. I can upload the pdf if you would like.

#### Attached Files:

• ###### Problem7Bonus.pdf
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4. Jun 23, 2014

### Staff: Mentor

Are you supposed to assume that the acceleration and initial speed of B is the same in both situations?

5. Jun 23, 2014

### antonisz

I believe so, since the problem says that both are a constant.

6. Jun 23, 2014

### Staff: Mentor

Good. That fact provides the 'extra' information needed to solve for them. (For each situation, B must travel a certain distance in a certain time.)

7. Jun 23, 2014

### BiGyElLoWhAt

Ok, well firstly, you substituted inconsistently to get your -4/3, if ta= some number, then tb should equal that same number.

Secondly, the fact that it says vib = constant makes me thing you're just supposed to treat it as an arbitrary constant, as you litterally cannot solve this problem numerically. Set vib = c or k or lambda or whatever you like, and use that for vib, and solve for ab in terms of your constant of choice.

8. Jun 23, 2014

### Staff: Mentor

Good.

You used the wrong distance.

9. Jun 23, 2014

### BiGyElLoWhAt

Ahhh, sorry about that, I didn't think about that, but at the same time, it's not stated in the problem (at least not very clearly i.m.o.).

10. Jun 23, 2014

### antonisz

Sorry that was a typo on my part.

What I had on paper was 160 = vib(3) + 1/2 (ab)(3)2

11. Jun 23, 2014

### antonisz

280 = vib(4) + 1/2 (ab)(4)2
160 = vib(3) + 1/2 (ab)(3)2

So I solved for ab and I got 11.85 / vib = ab

I plugged that into the 280 = vib(4) + 1/2 (ab)(4)2 equation and I got an answer of 0.34 m/s. Which still doesn't seem right.

But continuing on I plugged 0.34 m/s into 280 = vib(4) + 1/2 (ab)(4)2 to solve for ab and received an answer of 34.82 m/s2

I'm not sure if that's what you guys got for your answer or not.

But something still feels off. I really appreciate you guys and gals taking the time out of your busy summer day to teach me some physics...lol.

12. Jun 23, 2014

### BiGyElLoWhAt

Well, I didn't actually crunch the numbers, but you can always take situation a, and (knowing the time) plug in your vib and ab values and solve for the distance traveled, then check to see if that matches your starting point.

13. Jun 23, 2014

### BiGyElLoWhAt

*and then do the same for situation b.

14. Jun 23, 2014

### antonisz

I must have messed up somewhere. Thanks anyway.

15. Jun 23, 2014

### BiGyElLoWhAt

well here let me work through it and I'll see what I can find..

16. Jun 23, 2014

### antonisz

Thank you again, I hope I didn't make you feel like you have too lol. To be honest its probably something like a math error.

17. Jun 23, 2014

### BiGyElLoWhAt

That's fine, but I'm bored at work... so...

18. Jun 23, 2014

### Staff: Mentor

Your answers match my own. But I agree, it does feel off. Those numbers are not very realistic. (But it's probably meant as a simple exercise, not as a physically realistic situation.)

19. Jun 23, 2014

### antonisz

I could see my professor giving us a bonus question like that to be honest. His name is Doc Al as well...what a coincidence! Thank you for your confirmation, I needed those 6 points!