Find the position function x(t) of a moving particle with the given acceleration a(t), intial position x0 = x(0), and initial velocity v0 = v(0).

a(t) = 1/sqRoot(t+4), v0 = -1, x0 = 1

Im at a loss. Any help is appreciated.

Printable View

- Sep 2nd 2010, 07:50 PMjzelltPosition Function
Find the position function x(t) of a moving particle with the given acceleration a(t), intial position x0 = x(0), and initial velocity v0 = v(0).

a(t) = 1/sqRoot(t+4), v0 = -1, x0 = 1

Im at a loss. Any help is appreciated. - Sep 2nd 2010, 08:15 PMTKHunny
You are expected to know:

s(t) = x'(t)

a(t) = s'(t)

Given a(t), you must find the antiderivative and use the value provided to determine the coefficient. This gives v(t).

Given v(t), you must find the antiderivative and use the value provided to determine the coeficcient. This gives x(t). - Sep 3rd 2010, 08:30 PMjzellt
Thanks but I'm still a bit lost. Can someone show how this problem is done so I have an example to use while doing the others? Thanks for any help

- Sep 10th 2010, 05:31 PMTKHunny
You have shown no work at all. How can anyone help you improve what you are doing?

You have a(t). Find its antiderivative and determine s(t). - Sep 11th 2010, 02:42 AMAckbeet
Just as a clarification of post # 2, in order to be consistent with the notation of the OP:

v(t) = x'(t)

a(t) = v'(t).