# Deriving an equation for displacement and acceleration (given velocity)

1. Nov 18, 2008

### southernbelle

1. The problem statement, all variables and given/known data
For 0<t<1, v(t) = t +3
For 1<t<2, v(t) = 5-t
Assume x(o)=0

A) Draw corresponding displacement and acceleration diagrams.
B) Determine the equation for each segment

2. Relevant equations
Acceleration is the derivative of velocity.
Velocity is the derivative of displacement.

3. The attempt at a solution
I can draw the acceleration diagram and write the equation so no problem there.
My problem is drawing the displacement diagram.

I got the equations for displacement. They are:
For 0<t<1
t2/2 + 3t + C
For 1<t<2
5t - t2/2 + C

I cannot figure out how to evaluate that constant and plot that on a graph. Also, my teacher mentioned finding the area under the original curve and plotting that. The area = 3.5but that's not a point to plot. What do I dO?

2. Nov 18, 2008

### minger

It seems to me that you have everything except the constants of integration. You are told to assume that x(0) = 0, meaning that at time of zero, you have no displacement. Plug it in to get your constant.

3. Nov 18, 2008

### tiny-tim

Hi southernbelle!

x(0)=0, so t2/2 + 3t + C has to be 0 when t = 0, so C = … ?
ah … your 3.5 is just the area for t = 1 …

your teacher meant the area A(t) up to time t for any t

plot A(t) against t, and that's the displacement.

4. Nov 18, 2008

### HallsofIvy

Staff Emeritus
You have
For 0<t<1
x(t)= t2/2 + 3t + C
For 1<t<2
x(t)=5t - t2/2 + C
and x(0)= 0.

Be careful- the two "C"s are necessarily the same.

Use x(0)= 0 to find C in the first equation. Then use the fact that the two equations must give the same result at x= 1 to find C in the second equation.

5. Nov 18, 2008

### southernbelle

Okay, so

I evaluated the constants.
For the first equation:
C = 0
For the second equation
C = -1

But how do I plot those? Would I use the coordinates (0,0) and (1, -1) ?

I am thinking that the Constant is where you start on the y-axis and then you use the slope to go from there.

But the equation is not written in slope intercept form.
? :(

6. Nov 20, 2008

### minger

Well it's not slope intercept form because it's not a simple linear equation. Graphing these is quite easy. Time is your independant variable, it depends on nothing, so it's your x-axis. The velocity/disp/accel are dependent on time x = f(t), so it's your y-axis. Just start at t=0, plug it into your equation and put a point, then go to 0.1, or whatever you choose, and calculate x. Rinse and repeat until you get to time = 1.0 seconds, then switch to the other equation.