Question

a particle moves along the $x$-axis. the function $x(t)$ gives the particles position at any time $t \geq 0$
$x(t) = t^3 - 3t^2 + 7t - 6$
what is the particles acceleration $a(t)$ at $t = 3$
$a(3) = \square$

Explanation:

Step1: Find velocity $v(t)$ (1st derivative)

$v(t) = x'(t) = 3t^2 - 6t + 7$

Step2: Find acceleration $a(t)$ (2nd derivative)

$a(t) = v'(t) = 6t - 6$

Step3: Substitute $t=3$ into $a(t)$

$a(3) = 6(3) - 6$

Answer:

12