# Help with quick integration problem

1. Feb 11, 2004

### Pepsi24chevy

Say I got this velocity graph. Assuming X=0, at T=0, write correct algebraic expressions for x(t), v(t), and a(t) with appropriate numerical values inserted for all constants.

Graph: (connect lines(-) to make straight line going from 0,50 to 10, -50

Vx (m/s)
50
!-_
! - _
!_______________
! -_ 10
! -
! -_
-50
How would i go about solving this problem?
Well graph wont' show up right, but it starts at (0,50), and it a straight line down to (10,-50)

Last edited: Feb 11, 2004
2. Feb 11, 2004

### turin

This is unclear to me.

3. Feb 11, 2004

### Pepsi24chevy

Can u understand it now?

4. Feb 12, 2004

### HallsofIvy

The velocity graph is a straight line from (0,50) to (10,-50) so it should be easy to calculate that the slope is (-50-50)/(10-0)= -10.
The equation of the velocity is v= -10t+ 50.
Of course, the acceleration is just the slope of that line: -10.

I don't know what math you have to find the position function. If you have taken calculus, the position is an anti-derivative (integral) of the velocity function: the anti-derivative of v=-10t+50 is -5t2+ 50t+ C where C is some constant. Since we are told that x= 0 when t=0, x(0)= -5(0)+ 50(0)+ C= 0 so C= 0. The position function is x(t)= -5t2+ 50t.

If you can't use calculus, in this simple situation (acceleration is constant) you can argue that the change in position is the area under the graph. For fixed T>0, The area under the graph is a trapezoid with one base 50 (at t=0), the other base 50-10T (at t=T) and height t. The area of that trapezoid is (1/2)h(b1+b2)= (1/2)T(50+50-10T)=(1/2)(100T- 10T2)= 50T- 5T2. Since the initial x value was 0, the position is just that change in position:x(t)= 50t- 5t2 just as before.

Yet another way is to argue that, with constant acceleration, we can treat this a a constant velocity problem by averaging the first and last velocities: the first velocity is 50 and the last velocity (at t= T) is 50- 10T so the average is (50+ (50- 10T))/2 = 50- 5T. Moving for time T at constant velocity 50- 5T, the object will move distance T(50- 5T)= 50T- 5T2 just as before.

5. Feb 12, 2004

### paul11273

If I understand this question, you are given a graph with velocity on the y-axis and time on the x-axis.
You then have a straight line from (0,50) to (10,-50), right?

If so, then that line represents the velocity of an object at any given time. From this graph you can imagine a car that is initially traveling at 50mph at time=0, then slows down to zero mph at t=5, and begins to reverse direction to 50mph backwards.

Knowing the following relationships:
x(t) is position of the car at a any given time.
v(t) is velocity at any given time = x'(t)
a(t) is acceleration at any given time = x"(t) or v'(t).

From your graph, you can determine the slope of the velocity. Write this in the form of v(t)=mx+b.
To find the position, x(t), you integrate the expression you wrote for v(t).
To find the acceleration, a(t), you take the derivative of the v(t) expression you wrote.

I hope I understood the question, and let us know if this helps you.
Post your answer here and I will see if it agress with what I have, I hope so!!!

6. Feb 12, 2004

### paul11273

Halls of Ivy-
I agree with the method of your response, however I think you made an error on the slope. It should be -5. The start point of the graph is (50,0).
Recheck that and let me know what you think.