How to sketch a Position Graph --- Linear Alg/Diff EQs
Based off of the velocity graph (we are just given the velocity graph ----> no function whatsoever, just the graph), sketch the graph of the resulting position function x(t) for 0<=t<=10.
For some reason the position graph kind of looks like a cubic graph (answer is in the back of the book), but I have no idea how they got that.
I also see that (back of the book):
x(t) =
5/6t^2 if 0<=t<=3
5t-(15/2) if 3<=t<=7
1/6(-5t^2+100t-290) if 7<=t<=10
I'm not sure how they got those polynomials, although I can see how they got the intervals because on the velocity graph, with time (t) being on independent/x axis and velocity (v) being on the dependent/y axis; the graph goes uphill from (0,0) to (3,5). Then from (3,5) straight to (7,5). Then downhill from (7,5) to (10,0) ---- the graph makes a trapezoid shape.
Help anyone?
3 Comments • Newest first
Since you mentioned that this is for a differential equations course, I'm going to assume knowledge about integration of polynomials. The position function is the cumulative area under the velocity graph. That is, for every interval of constant increase/decrease in velocity, you integrate the velocity function present for that domain.
Note that v=(dx/dt).
Ex. For 0<=t<=3
The velocity function is a straight line with a positive slope of 5/3. This tells us that the velocity is in the form v=(5/3)t+c, where c=0 because we can sub in (v,t)=(0,0).
Thus the velocity function is (dx/dt)=(5/3)t.
We integrate both sides of this equation with respect to t, and we arrive at:
x=(5/6)t^2.
For the next two intervals, remember that your position function is the cumulative area, so you must take the distance travelled in any previous interval(s) into consideration.
Isn't this type of question a typical Calculus AB question?
post a picture of the graph