University:
Rice University
Course:
MATH 212 | Multivariable Calculus
Academic year:

2023
Views:

208

Pages:

1
Author:

Deskov

Particle Motion. Example Problem 1 Let's look at a fairly straightforward position-velocity-acceleration question. We are starting from a position of mass on a spring. Using the equation for spring motion, we can find the velocity for this system and then evaluate the acceleration by taking the derivative of velocity. We want to label x prime of t as equal to v of t so that we give the impression of being more careful and deliberate in our work. So, of cosine, that is negative sine. Thus, -6 sin(t) inside the function is equal to 2 pi t times – pi/2. The derivative of the inside function is 2, so – 6 sin(t) = 2 pi t times – pi/2 multiplied by 2. I hope that you are finding that you are improving in your understanding of the chain rule. The velocity at time 1 would be -12 times the sine of 2 pi times 1. This is 2 pi, minus pi over two is 3/2. The sine of 3 pi over 2 is -1, so the velocity at time 1 would be 12. Note that I did not just say "v of t equals 12" here. That would be incorrect because v of 1 equals 12, so we have to keep the two equalities separate. That's what v of 1 equals. It is something that the AP test looks for, like a continuity issue of sorts. We'll take the derivative of position with respect to time and call it velocity. The second derivative of position is called acceleration, so we'll call that a of t. The derivative of -12 sine is -12 cosine. The chain rule shows that the derivative of 2 times the variable t is equal to 2 times the derivative of t plus the constant pi divided by 2. And then the acceleration at time 1 would be -24 times the cosine of 3 pi over 2, multiplied by two. The cosine of 3 pi over 2 is zero; therefore the acceleration is zero. And you have your answer.