# 2 vehicles in kinematics

1. Feb 1, 2015

### Bry2323

A truck and a car both start from rest at the same time on a straight track. The car is 25.0 m behind the truck at the start. The truck accelerates at 3.70 m/s^2 and the car accelerates at 4.40 m/s^2. How long does it take for the car to catch up to the truck?

Hey all, first post, didn't think it'd be so quick in this college physics class! Can someone help me figure this out...I have an idea that displacement of truck should equal the car plus 25.0 m right? Just having trouble getting there. Thank you!

2. Feb 1, 2015

### haruspex

Please don't delete the template. It's there to be filled in. What relevant equations do you know? Please show some attempt.

3. Feb 1, 2015

### Bry2323

Sorry, didn't realize that until i searched around a little more.

So far I have displacement of truck equals 1/2a x t^2 = 1.85 m/s^2(t^2)
I am thinking displacement of car = displacement of truck + 25.0 m = 1/2at^2

I am left with 1.85 m/s^2(t^2) = 2.20 m/s^2(t^2) + 25.0 m?? This is where I am so very lost.

Thank you again.

4. Feb 1, 2015

### haruspex

Which starts in front? What are the two displacements at t=0?

5. Feb 1, 2015

### Bry2323

The truck starts in front. The displacement for the car is 0m and truck is 25m?

6. Feb 1, 2015

### haruspex

Right, so what is your equation now?

7. Feb 1, 2015

### Bry2323

I'm not sure, are you pointing out that displacement of truck should be displacement of car + 25.0 m?

8. Feb 1, 2015

### haruspex

Yes. What you had before gives the wrong result at t=0.

9. Feb 1, 2015

### Bry2323

I think I had it correct underneath that? I really don't know how to simplify that last line if that is the correct formula

10. Feb 1, 2015

### haruspex

Your formula has the form At2+B=Ct2. Can you not see how to simplify that?

11. Feb 1, 2015

### Bry2323

I really can't, is that a quadratic? It's been awhile since I have been in Algebra.

12. Feb 1, 2015

### Bry2323

ok I think i got it but I think i've done way too many steps

2.20 m/s^2(t^2) +25.0m = 1.85 m/s ^s (t^2)

.35 m/s^2(t^2) +25.0m = 0

.35(t^2) = -25.0s^2

t^2 = -71.43 s^2

t= 8.45 s

13. Feb 1, 2015

### haruspex

The left hand side has the truck's acceleration, so presumably represents the truck's displacement. So your equation says displacement of truck = displacement of car + 25m.
This is why you get an impossibility here:
t2 cannot be negative. It is not valid simply to ignore the minus sign because it's inconvenient. In the present case, ignoring it does happen to give the right answer, because the minus sign should not have been there in the first place.

14. Feb 1, 2015

### Bry2323

Ok well how do i move the 25.0 m to the right side without it being negative?

15. Feb 1, 2015

### haruspex

This equation is wrong:
It does not correspond to your (correct) observation that "displacement of car = displacement of truck + 25.0 m". It has reversed it.
Correct the equation and the minus sign will go away.

16. Feb 1, 2015

### Bry2323

ok perfect thank you, so what exactly am i saying when i say the displacement of car = displacement of truck +25.0 m, my brain is fried after today but im having trouble understanding the theory behind it in lamens terms, my mind is wanting to think its the car +25 to make up for the deficit at the beginning but i know thats not right.

17. Feb 1, 2015

### haruspex

First, you need to make clear what you mean by that. It is correct if you mean their final displacements from their respective starting positions.
For each, its displacement from its starting position at time t is acceleration*t2/2.
So you have:
displacementcar = displacementtruck+25m
accelerationcar*t2/2 = accelerationtruck*t2/2+25m

18. Feb 1, 2015

### Bry2323

I think im confusing myselfs because im thinking the only way the addition of 25m would need to be added was if the accelerations were the same, not sure why i cant grasp the idea, think im just burnt out

19. Feb 1, 2015

### haruspex

I think your confusion comes from not being clear in what you meant by displacement in "displacement of car = displacement of truck + 25.0 m".
If it means their respective displacements from their respective starting positions when they meet, it is true. If it means respective displacements from a common origin when they start, it is backwards.

20. Feb 1, 2015

### Bry2323

Ok, I'll take a break and think it over, just trying to make sure I truly understand everything, you've been a big help, cheers!