Thursday, November 09, 2006

Ch. 3 Test Question #16

A particle moves along the x-axis, its position at time t given by x(t) = -(t^4) + 4(t^3) - 2, where t is measured in seconds and x in meters.

#16 When is the particle speeding up?

First, find the velocity by taking the derivative of the position function. You get:
v(t) = -4(t^3) + 12(t^2)

Then, find the acceleration by taking the derivative of the velocity [this is the second derivative of the position function]
a(t) = -12(t^2) + 24t

graph the velocity and acceleration functions using a reasonable window. (I apologize for not being able to show you a picture, but my internet is unusually slow as of now)

the particle is speeding up when the values of the velocity and acceleration functions HAVE THE SAME SIGN. they have the same sign when t is greater than 0 or less than 2 [both positive] and when t is greater than 3 [both negative]. (for some odd reason, because of html tags, blogger will not let me actually put in the greater than and less than signs.)

an alternative method (not mentioned in the book, I believe but correct me if I'm wrong) would be to sketch the graph, and take the absolute values of the graph of the velocity funcition (flip the negative parts by reflecting it over the x-axis). then if the graph is going downwards, it is slowing down. if it is going upwards, it is speeding up.


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