Solving Velocity Time Graph: Car 1 vs. Car 2

[Nirupt]

Homework Statement

http://imageshack.us/photo/my-images/96/caracarbgraph2.png/[PLAIN]


[h2]Homework Equations[/h2]
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.


[h2]The Attempt at a Solution[/h2]
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.

 

[rollingstein]

http://img96.imageshack.us/img96/7369/caracarbgraph2.png

 

[ehild]

Homework Statement

http://imageshack.us/photo/my-images/96/caracarbgraph2.png

Homework Equations

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.

The Attempt at a Solution

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.

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

ehild

 

[rollingstein]

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

My bad.

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

 

[ehild]

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

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

ehild

 

[rollingstein]

That is true when they start from the same place.

ehild

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

 

[Nirupt]

Why -14?

 

[rollingstein]

Why -14?

What's acceleration?

 

[ehild]

That is not -14. Check.

ehild

 

[rollingstein]

That is not -14. Check.

ehild

Right. My blunder. -6.

 

[rollingstein]

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

 

[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.

 

[rollingstein]

It's a quadratic right? How did you know to use that formula?

Which?

 

[ehild]

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.

What kind of motions are involved?

ehild

 

[Nirupt]

Displacement, acceleration, initial velocity, and time?

 

[ehild]

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

ehild

 

[Nirupt]

x =v₀t + ½*at²

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

 

[ehild]

If the set-up was that both cars started from the same position x=0, then the position of car A is x_A=70t-3t² as function of the time t and the position of car B is x_B=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, x_A=X_B. 40t=70t-3 t²

ehild