Question

suppose that the position of a particle is given by $s = f(t) = 4t^3 + 5t + 9$.
(a) find the velocity at time $t$.
$v(t) = \frac{\mathrm{m}}{\mathrm{s}}$
(b) find the velocity at time $t = 3$ seconds.
$\frac{\mathrm{m}}{\mathrm{s}}$
(c) find the acceleration at time $t$.
$a(t) = \frac{\mathrm{m}}{\mathrm{s}^2}$
(d) find the acceleration at time $t = 3$ seconds.
$\frac{\mathrm{m}}{\mathrm{s}^2}$
question help: video

Explanation:

Step1: Differentiate position for velocity

The position function is $s(t)=4t^3 + 5t + 9$. Using the power rule $\frac{d}{dt}(t^n)=nt^{n-1}$, we get:
$$v(t)=\frac{d}{dt}(4t^3 + 5t + 9)=12t^2 + 5$$

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

Calculate the velocity at $t=3$:
$$v(3)=12(3)^2 + 5=12\times9 + 5=108 + 5=113$$

Step3: Differentiate velocity for acceleration

Differentiate $v(t)=12t^2 + 5$ using the power rule:
$$a(t)=\frac{d}{dt}(12t^2 + 5)=24t$$

Step4: Substitute $t=3$ into $a(t)$

Calculate the acceleration at $t=3$:
$$a(3)=24\times3=72$$

Answer:

(a) $12t^2 + 5$
(b) $113$
(c) $24t$
(d) $72$