Question

suppose the position of a particle, s(t) is given by s(t)=t^2 - 6t + 23 feet, where t is time in seconds. (a) find the velocity function at time t. s(t)=v(t)= (b) what is the velocity after 3 seconds with correct units? (c) find all value(s) of t (with correct units) for which the particle is at rest (i.e., where v(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 position function

The derivative of $s(t)=t^{2}-6t + 23$ using the power - rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$ gives $v(t)=s'(t)=2t-6$.

Step2: Evaluate velocity at $t = 3$

Substitute $t = 3$ into $v(t)$: $v(3)=2\times3-6=0$ feet per second.

Step3: Find when particle is at rest

Set $v(t)=0$. So, $2t-6 = 0$. Add 6 to both sides: $2t=6$. Then divide by 2: $t = 3$ seconds.

Answer:

(a) $2t - 6$
(b) 0 feet per second
(c) 3 seconds