Question

suppose the position of a particle, s(t) is given by s(t)=t^{2}-5t + 22 feet, where t is time in seconds. (a) find the velocity function at time t. s(t)=v(t)= (b) find s(3), the velocity after 3 seconds. include units. (c) find all value(s) of t (with correct units) for which the particle is at rest (i.e., where s(t)=0). if there are no such values, enter none. if there is more than one value, list them separated by commas. t=

Explanation:

Step1: Differentiate to find velocity function

The derivative of \(s(t)=t^{2}-5t + 22\) using the power - rule \((x^n)^\prime=nx^{n - 1}\).
\[s^\prime(t)=v(t)=\frac{d}{dt}(t^{2}-5t + 22)=2t-5\]

Step2: Evaluate \(s^\prime(3)\)

Substitute \(t = 3\) into \(v(t)=2t - 5\).
\[v(3)=2\times3-5=6 - 5=1\]
The units of velocity are feet per second since \(s\) is in feet and \(t\) is in seconds. So \(s^\prime(3)=1\) feet per second.

Step3: Find when the particle is at rest

Set \(s^\prime(t)=0\), so \(2t-5 = 0\).
Solve for \(t\):
\[2t=5\Rightarrow t=\frac{5}{2}\text{ seconds}\]

Answer:

(a) \(2t - 5\)
(b) \(1\) feet per second
(c) \(\frac{5}{2}\text{ seconds}\)