suppose the position of a particle, s(t) is given by s(t)=t² - 12t + 27 feet, where t is time in…

suppose the position of a particle, s(t) is given by s(t)=t² - 12t + 27 feet, where t is time in seconds.\n(a) find the velocity function at time t.\n s(t)=v(t)=\n(b) find s(3), the velocity after 3 seconds. include units.\n(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.\n t=

suppose the position of a particle, s(t) is given by s(t)=t² - 12t + 27 feet, where t is time in seconds.\n(a) find the velocity function at time t.\n s(t)=v(t)=\n(b) find s(3), the velocity after 3 seconds. include units.\n(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.\n t=

Answer

Explanation:

Step1: Differentiate to find velocity

The derivative of $s(t)=t^{2}-12t + 27$ using the power - rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$. $v(t)=s'(t)=\frac{d}{dt}(t^{2}-12t + 27)=2t-12$

Step2: Evaluate velocity at $t = 3$

Substitute $t = 3$ into $v(t)$. $v(3)=2\times3-12=6 - 12=-6$ feet per second

Step3: Find when particle is at rest

Set $v(t)=0$. $2t-12 = 0$ Add 12 to both sides: $2t=12$ Divide both sides by 2: $t = 6$ seconds

Answer:

(a) $2t-12$ (b) $-6$ feet per second (c) $6$ seconds