Question

1. suppose that a particle has position ( s(t) ) feet at time ( t ) seconds and a velocity function ( s^{prime}(t)=3 sin (t) mathrm{ft} / mathrm{s} ). find the displacement (change in position) from time ( t = 0 ) seconds to time ( t=pi ) seconds.

Explanation:

Step1: Recall the displacement formula

The displacement from \(t = a\) to \(t = b\) is given by \(\int_{a}^{b}v(t)dt\), where \(v(t)\) is the velocity function. Here, \(a = 0\), \(b=\pi\), and \(v(t)=s^{\prime}(t)=3\sin(t)\). So, the displacement \(D=\int_{0}^{\pi}3\sin(t)dt\).

Step2: Integrate the function

We know that \(\int\sin(t)dt=-\cos(t)+C\). Using the fundamental theorem of calculus \(\int_{0}^{\pi}3\sin(t)dt=3[-\cos(t)]_{0}^{\pi}\).

Step3: Evaluate the definite - integral

First, substitute \(t = \pi\) and \(t = 0\) into \(-\cos(t)\).
When \(t=\pi\), \(-\cos(\pi)=-(- 1)=1\). When \(t = 0\), \(-\cos(0)=-(1)=-1\).
Then \(3[-\cos(t)]_{0}^{\pi}=3(1-(-1))\).

Step4: Simplify the expression

\(3(1 + 1)=6\).

Answer:

The displacement of the particle from \(t = 0\) to \(t=\pi\) is \(6\) feet.