Answer
$\tau = (48t~\hat{k})~N\cdot m$
Work Step by Step
$r = 4.0t^2~\hat{i}-(2.0t+6.0t^2)~\hat{j}$
$v = \frac{dr}{dt} = 8.0t~\hat{i}-(2.0+12t)~\hat{j}$
$a = \frac{dv}{dt} = 8.0~\hat{i}-12~\hat{j}$
We can find an expression for the force on the particle:
$F = ma$
$F = (3.0)(8.0~\hat{i}-12~\hat{j})$
$F = 24~\hat{i}-36~\hat{j}$
We can find an expression for the torque:
$\tau = r \times F$
$\tau = [4.0t^2~\hat{i}-(2.0t+6.0t^2)~\hat{j}] \times (24~\hat{i}-36~\hat{j})$
$\tau = (48t+144t^2-144t^2)~\hat{k}$
$\tau = (48t~\hat{k})~N\cdot m$
