Question

a particle moves along the $x$-axis. the function $x(t)$ gives the particles position at any time $t \geq 0$

$x(t) = t^4 - 9t^3 + 2t - 1$

what is the particles acceleration $a(t)$ at $t = 4$
$a(4) = \square$

Explanation:

Step1: Find velocity function $v(t)$

Velocity is the first derivative of position:
$v(t) = x'(t) = 4t^3 - 27t^2 + 2$

Step2: Find acceleration function $a(t)$

Acceleration is the derivative of velocity:
$a(t) = v'(t) = 12t^2 - 54t$

Step3: Evaluate $a(t)$ at $t=4$

Substitute $t=4$ into $a(t)$:
$a(4) = 12(4)^2 - 54(4)$
$= 12(16) - 216$
$= 192 - 216$

Answer:

$-24$