Find Velocity and Time from an Acceleration vs Time Graph

[CGI]

Homework Statement

The rocket car is subjected to a constant acceleration of Ac = 6 m/s^2 until T1 = 15 s.
The brakes are then applied which causes a deceleration at the rate shown until the car stops. Determine the maximum speed of the car and the time T when the car stops.

Homework Equations

Area of Rectangle = L x W
Area of a Triangle = 1/2(b)(h)

The Attempt at a Solution

So I know that the area under the curve for an acceleration vs time graph would represent the velocity, but without having what the final time T was or what the final acceleration was, I'm not sure how to go about it. I know that at 15 seconds, the velocity is 90 m/s and that's about it.

Any help would really be appreciated!

 

[TSny]

Are you asking about the first part of the question: getting the maximum speed? If so, at what time does maximum speed occur? Before the brakes are applied, after the brakes are applied, or at the instant the brakes are applied?

 

[CGI]

I would say before the brakes, so that must mean that the 90 m/s IS the maximum speed that car attains.

Is that correct?

 

[TSny]

Yes, the car is picking up speed all the way until the brakes are applied. So, maximum speed occurs at t = 15 s.

 

[CGI]

Okay that just leaves me with finding the time at which it stops. So that means the velocity should equal zero which means that the area of the triangle should also be equal to 90 which I think might help, how exactly would I go about finding the time?

 

[TSny]

So that means the velocity should equal zero which means that the area of the triangle should also be equal to 90

Good!

how exactly would I go about finding the time?

Note the little triangle with sides of ratio 1 to 2.

 

[CGI]

Oh I forgot to ask about that! Unfortunately, I'm not entirely sure about what that means..

 

[TSny]

Oh I forgot to ask about that! Unfortunately, I'm not entirely sure about what that means..

It allows you to get the slope of this part of the acceleration graph.

 

[CGI]

Okay, so the slope is rise over run, which would mean that the slope here is 1/2 right? But I'm not sure how I would use that to get the time

 

[TSny]

Try to write an expression for the area of the triangle of the graph in terms of the time interval Δt (between t₁ and t).

 

[CGI]

Hmmm.. it seems like the only thing I can think of is the integral from t1 to t of the area of the triangle is equal to the velocity. Is that in the right direction?

 

[TSny]

Hmmm.. it seems like the only thing I can think of is the integral from t1 to t of the area of the triangle is equal to the velocity. Is that in the right direction?

Yes. And the integral is just the area of the triangle. You are ultimately looking for the base of that triangle. Can you express the height of the triangle in terms of the base? You can use similar triangles with one triangle being the little triangle with sides 1 and 2.

 

[CGI]

Would the height just be equal to half the base?

 

[TSny]

Yes.

 

[CGI]

Okay, so the integral would be:

90 = integral of b^2/4dt from 15 to t? Which becomes b^3/12t?

 

[TSny]

Okay, so the integral would be:

90 = integral of b^2/4dt from 15 to t? Which becomes b^3/12t?

b^2/4dt is the integral (i.e., the area of the triangle).

 

[CGI]

Okay so something more like this?

90 = (t^2 - 225)/4 in which case, t = 24.2?

 

[TSny]

I should have said that b²/4 is the area (instead of b²/4 dt). You know what this area must equal. So, what do you get for the base b?

 

[CGI]

I get 18.97 for the b. Which would be the answer for t right?

 

[TSny]

I agree with your answer for b (assuming you're still using the second as a unit of time). But I don't agree with your answer for the instant of time t when the car comes to rest.

 

[CGI]

Wait! But that's just the base of the triangle..
Wouldn't t be the 15 + 18.97?

 

[TSny]

Yes. Don't forget the units.

 

[CGI]

Wow! Thank you so much! I don't understand why that took me so long, but I appreciate all the help today!

 

[TSny]

OK. Good work.