Question

question 3
a particle is moving with the given data. find the position of the particle.
$v(t)=\sin t - \cos t$, $s(0)=4$
$\bigcirc s(t)=5 - \cos t - \sin t$
$\bigcirc s(t)=1 - t - \sin t$
$\bigcirc s(t)=\cos^2 t$
$\bigcirc s(t)=\cos t - \sin t$
$\bigcirc s(t)=1 - \cos t + \sin t$

Explanation:

Step1: Integrate velocity function

Since \(v(t)=\sin t-\cos t\), and \(s(t)=\int v(t)dt\).
Using \(\int\sin tdt = -\cos t + C_1\) and \(\int\cos tdt=\sin t + C_2\), we have \(s(t)=-\cos t-\sin t + C\).

Step2: Use initial condition

Given \(s(0) = 4\). Substitute \(t = 0\) into \(s(t)\):
\(s(0)=-\cos(0)-\sin(0)+C\).
Since \(\cos(0) = 1\) and \(\sin(0)=0\), then \(4=-1 - 0+C\).
Solving for \(C\), we get \(C = 5\).

Answer:

\(s(t)=5-\cos t-\sin t\)