Homework Help: Velocity time graph

1. Feb 3, 2013

Nirupt

1. The problem statement, all variables and given/known data
http://imageshack.us/photo/my-images/96/caracarbgraph2.png/[PLAIN]

[b]2. Relevant equations[/b]
This image is 2 cars, and Car 1 is starting at a higher velocity rate than Car 2, which is constant @ t = 0. Now then, the car is ahead and then decreases speed over time, I am trying to figure out when car 2 well eventually catch up to car 1.

[b]3. The attempt at a solution[/b]
I am not sure if there is any equation to this, but it looks like they go the same speed at 5 seconds, the slope of car 1 is 6 m/s^2 (70 m/s - 10 m/s = 60 m/s/10s = 6m/s^2) However I'm not even sure if that actually helps.. since car 2 has a slope of 0.

2. Feb 3, 2013

rollingstein

http://img96.imageshack.us/img96/7369/caracarbgraph2.png [Broken]

Last edited by a moderator: May 6, 2017
3. Feb 3, 2013

ehild

You need to know what was the initial distance between the cars.

ehild

4. Feb 3, 2013

rollingstein

[STRIKE]40 t = 70 t + 1/2 * (-14) t^2[/STRIKE]

40 t = 70 t + 1/2 * (-6) t^2

Last edited: Feb 3, 2013
5. Feb 3, 2013

ehild

That would be true when they start from the same place (if you used the correct acceleration).

ehild

6. Feb 3, 2013

rollingstein

In the absence of other info. I assumed that, yes.

7. Feb 3, 2013

Why -14?

8. Feb 3, 2013

rollingstein

What's acceleration?

9. Feb 3, 2013

ehild

That is not -14. Check.

ehild

10. Feb 3, 2013

rollingstein

Right. My blunder. -6.

11. Feb 3, 2013

rollingstein

40 t = 70 t + 1/2 * (-6) t^2

12. Feb 3, 2013

Nirupt

It's a quadratic right? How did you know to use that formula?
so A = -3, B = 30, and C= 0
= -5 (+/-) SQRT(1) = 0, and 10

So 10 seconds is the answer.

13. Feb 3, 2013

rollingstein

Which?

14. Feb 3, 2013

ehild

What kind of motions are involved?

ehild

15. Feb 3, 2013

Nirupt

Displacement, acceleration, initial velocity, and time?

16. Feb 3, 2013

ehild

Car A moves with constant acceleration. You should know the formula for displacement in terms of time.

ehild

17. Feb 3, 2013

Nirupt

x =v0t + ½*at2

But it was setup that x was 40t because the distance was needed between the 2 cars correct?

18. Feb 4, 2013

ehild

If the set-up was that both cars started from the same position x=0, then the position of car A is xA=70t-3t2 as function of the time t and the position of car B is xB=40t. Car A leaves car B at the start, but slows down and car B catches it at the end. At that instant, both cars are at the same place again, xA=XB. 40t=70t-3 t2

ehild