[maths] Re: position function; calculus
- From: "Ned Granic" <ngranic@xxxxxxx>
- To: <maths@xxxxxxxxxxxxx>
- Date: Thu, 7 Dec 2006 04:54:15 -0700
Already Nelson! Thanks a lot. What I've got then is:
s(t) =-cos t-sin t+6t+1.
Correct?
----- Original Message -----
From: Nelson Blachman
To: maths@xxxxxxxxxxxxx
Sent: Thursday, December 07, 2006 1:48 AM
Subject: [maths] Re: position function; calculus
Ned,
Your integrations are correct, but your substitutions of the values at time
t=0 are wrong.
While sin 0 = 0, cos 0 is not 0.
HTH
--Nelson
----- Original Message -----
From: Ned Granic
To: maths@xxxxxxxxxxxxx
Sent: Tuesday, December 05, 2006 10:45 AM
Subject: [maths] position function; calculus
Hi all,
Here is one little confusing question that needs some clarification.
A particle is moving with the given data.
Find the position of the particle.
a(t) = cos t + sin t -- acceleration
s(0) = 0 -- position at 0 seconds
v(0) = 5 -- initial velocity.
a(t) = v^'(t) -- acceleration is a derivative of a velocity
= cos t + sin t, that is, velocity is antiderivative of the acceleration,
therefore, the velocity function is:
v(t) = sin t-cos t+C
which means that:
5 = 0+C, that is:
C = 5, and the required velocity function is:
v(t) = sin t-cos t+5.
Then, the velocity function is a derivative of the position function:
v(t) = s^'(t), that is, the position function is antiderivative of the
velocity function:
s(t) = -cos t-sin t+5t+D, that is:
0 = 0+D, or:
D = 0, so the required position function here is:
s(t) = -cos t-sin t+5t.
The question is, what is the position of the particle with the given data?
Many thanks in advance!
Ned
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