# Homework Help: Comparison of velocity at the end of distance

1. Mar 23, 2015

### gijungkim

1. The problem statement, all variables and given/known data
Carts A and B have equal masses and travel equal distances D on side-by-side straight frictionless tracks while a constant force F acts on A and a constant force 2F acts on B. Both carts start from rest. The velocities vA and vB of the bodies at the end of distance D are related by

a.vB = vA.

b.vB = vA.

c.vB = 2 vA.

d.vB = 4 vA.

e.vA = 2vB.

2. Relevant equations
F = ma
x = vot + 0.5 at^2
Vf = Vo + at

3. The attempt at a solution
Answer is B but I have no clue why
First, I just picked a random number to make it easier. I picked acceleration of 2m/s^2 for A and 4m/s^2 for B (since it has twice more force than A)
and I picked x as 1m
Then I'll get
1 = 0 + 0.5 * 2 * t^2, t = 1 for A
1 = 0 + 0.5 * 4 * t^2, t = 0.707 for B
If I plug this in to Vf = V0 + at
Vf = 2 for A
Vf = 2.83 for B...
To me vB = vA doesn't make sense.. Can anyone help me

2. Mar 23, 2015

### TSny

Was there any information given regarding the directions of the forces?

3. Mar 23, 2015

### PeroK

Answers a. and b. are the same. Are you sure b. isn't something else?

Your calculations are correct. Can you see a way to generalise them to get the answer?

4. Mar 23, 2015

### gijungkim

That is exactly what question stated. I copied and pasted it.

5. Mar 23, 2015

### gijungkim

Probably the answer is wrong haha I'm just using random questions from the internet to practice so.

6. Mar 23, 2015

### PeroK

It would be good practice to get the answer for this without multiple choice.

You could generalise the calculations you've already done. Or you could consider energy.

7. Mar 23, 2015

### gijungkim

Is the answer V2 = sqrt 2 * V1 ???

8. Mar 23, 2015

### PeroK

It is!

9. Mar 23, 2015

### gijungkim

Thank you!! :)