Question

1. - / 4.54 points

the equation of motion of a particle is $s = t^3 - 12t$, where $s$ is in meters and $t$ is in seconds. (assume $t \geq 0$.)
(a) find the velocity and acceleration as functions of $t$.
$v(t) = \square$
$a(t) = \square$
(b) find the acceleration after 8 s.
$\square$ $\text{m/s}^2$
(c) find the acceleration when the velocity is 0.
$\square$ $\text{m/s}^2$
resources
read it watch it

Explanation:

Step1: Find velocity function

Velocity is derivative of position: $v(t) = s'(t) = 3t^2 - 12$

Step2: Find acceleration function

Acceleration is derivative of velocity: $a(t) = v'(t) = 6t$

Step3: Calculate acceleration at t=8

Substitute t=8 into $a(t)$: $a(8) = 6×8 = 48$

Step4: Find t when velocity is 0

Set $v(t)=0$: $3t^2 -12=0 → t^2=4 → t=2$ (t≥0)

Step5: Acceleration at t=2

Substitute t=2 into $a(t)$: $a(2)=6×2=12$

Answer:

(a) $v(t)=3t^2 - 12$, $a(t)=6t$
(b) 48
(c) 12