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