# Homework Help: Solving for time with an non consistant acceleration

1. Aug 4, 2013

### laketri

Problem
Several students are riding in bumper cars at an amusement park. The mass of car A and its occupants is 250kg. The combined mass of car B and its occupants is 200kg. Car A is 15m away away from car B and moving to the right at 2.0m/s, as shown, when the driver decides to bump into car B, which is at rest.

(A) Car A accelerates at 1.5m/s2 to a speed of 5.0m/s and then continues at a constant velocity until it strikes car B. Calculate time for car A to travel the 15m.

Attmept
I assumed that you would use v=d/t until I realized that velocity isn't constant. So I looked to the kinematic equations, and thought i would use d=vi*t+1/2*a*t2 until i noticed that acceleration isn't constant either.

I'm going into AP Physics B after I took Honors Physics last year. I've never had to work with non constant acceleration before and I don't know what to do with it. I've looked online and never found anything relevant to what I need.

I don't want an answer to the question, I just need to know what I use to get to the answer.

2. Aug 4, 2013

### CWatters

You didn't post the diagram.

Can you break it down into two parts. eg The time spent under constant acceleration and the time spent at constant velocity?

3. Aug 4, 2013

### laketri

I didn't paste the diagram because it is an actual paper and I don't have a scanner, so there's no way for me to, sorry.

I can't break down the time because that's what I'm solving for, unless I don't understand you correctly

4. Aug 4, 2013

### SteamKing

Staff Emeritus
You should take CWatters advice. Let's catalog what you know about Car A:

Mass of Car A + occupants = 250 kg
Distance Car A to Car B = 15 m

Initial velocity of Car A = 2.0 m/s
Final velocity of Car A = 5.0 m/s

Car A accelerates from 2.0 m/s to 5.0 m/s with acceleration of 1.5 m/s^2
How long does this take?
How far does Car A travel while accelerating?
Is this distance < 15 m?
If yes, then figure out how long it takes Car A to from this point until it is 15 m from the starting point, assuming Car A is moving at a constant velocity of 5.0 m/s.

5. Aug 4, 2013

### laketri

So, if i did this right, it would take 2 seconds to accelerate to 5.0m/s
so using d=vi*t+1/2*a*(t2)
I could find that the distance is 7 m
Then using simple algebra, take 7m away from 15m, then deviding by 5 to get the amount of seconds traveled in 5m/s, I could find the time to be 3.6

6. Aug 4, 2013

### haruspex

All correct, except for the 3.6 at the end.

7. Aug 4, 2013

### laketri

I was adding the previous 2 seconds to the 1.6 which was received from 15-7/5

8. Aug 4, 2013

OK, good.