Deriving an equation for displacement and acceleration (given velocity)

AI Thread Summary
The discussion focuses on deriving equations for displacement and acceleration based on given velocity functions for two time intervals. For the first interval (0<t<1), the displacement equation is t²/2 + 3t + C, and for the second interval (1<t<2), it is 5t - t²/2 + C. The constant C can be determined using the initial condition x(0)=0, resulting in C=0 for the first equation and C=-1 for the second. Participants emphasize the importance of plotting the area under the velocity curve to visualize displacement over time. The conversation concludes with guidance on how to graph these equations effectively.
southernbelle
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Homework Statement


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


Homework Equations


Acceleration is the derivative of velocity.
Velocity is the derivative of displacement.


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?
 
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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 into get your constant.
 
southernbelle said:
For 0<t<1, v(t) = t +3
For 1<t<2, v(t) = 5-t
Assume x(o)=0

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.

Hi southernbelle! :smile:

x(0)=0, so t2/2 + 3t + C has to be 0 when t = 0, so C = … ? :wink:
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?

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. :smile:
 
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.
 
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.
? :(
 
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.
 
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