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suppose the position of a particle, s(t) is given by s(t)=t^2 - 11t + 1…

Question

suppose the position of a particle, s(t) is given by s(t)=t^2 - 11t + 19, 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.

Explanation:

Step1: Find the velocity function

The velocity function $v(t)$ is the derivative of the position - function $s(t)$. Given $s(t)=t^{2}-11t + 19$, using the power rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v(t)=s^{\prime}(t)=2t-11$.

Step2: Find the velocity at $t = 3$

Substitute $t = 3$ into the velocity function $v(t)$. So $v(3)=2\times3-11=6 - 11=-5$ feet per second.

Step3: Find when the particle is at rest

Set $v(t)=0$. So $2t-11 = 0$. Add 11 to both sides: $2t=11$. Then divide by 2: $t=\frac{11}{2}=5.5$ seconds.

Answer:

(a) $v(t)=2t - 11$
(b) $-5$ feet per second
(c) $5.5$ seconds