Question

suppose that the position of a particle is given by s = f(t) = 3t³ + 2t + 9. (a) find the velocity at time t. v(t) = m/s (b) find the velocity at time t = 3 seconds. m/s (c) find the acceleration at time t. a(t) = m/s² (d) find the acceleration at time t = 3 seconds. m/s²

Explanation:

Step1: Recall velocity - position relation

The velocity $v(t)$ is the derivative of the position function $s(t)$. Given $s(t)=3t^{3}+2t + 9$, using the power - rule for differentiation $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v(t)=\frac{d}{dt}(3t^{3}+2t + 9)$.
$v(t)=3\times3t^{2}+2\times1+0 = 9t^{2}+2$.

Step2: Calculate velocity at $t = 3$

Substitute $t = 3$ into the velocity function $v(t)$.
$v(3)=9\times3^{2}+2=9\times9 + 2=81 + 2=83$.

Step3: Recall acceleration - velocity relation

The acceleration $a(t)$ is the derivative of the velocity function $v(t)$. Since $v(t)=9t^{2}+2$, then $a(t)=\frac{d}{dt}(9t^{2}+2)$.
Using the power - rule, $a(t)=9\times2t+0 = 18t$.

Step4: Calculate acceleration at $t = 3$

Substitute $t = 3$ into the acceleration function $a(t)$.
$a(3)=18\times3=54$.

Answer:

(a) $9t^{2}+2$
(b) $83$
(c) $18t$
(d) $54$