Question

suppose the position of a particle, $s(t)$ is given by $s(t)=t^{2}-4t+35$ 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 position function

The velocity function is the derivative of the position function. Use the power rule $\frac{d}{dt}(t^n)=nt^{n-1}$:
$$v(t)=s'(t)=\frac{d}{dt}(t^2 - 4t + 35)=2t - 4$$

Step2: Substitute $t=3$ into $v(t)$

Calculate velocity at 3 seconds:
$$v(3)=2(3)-4=6-4=2$$

Step3: Solve $v(t)=0$ for $t$

Set velocity to 0 to find when the particle is at rest:
$$2t - 4=0 \implies 2t=4 \implies t=2$$

Answer:

(a) $v(t)=2t-4$
(b) 2 feet per second
(c) $t=2$ seconds