Answer
$d=204$ m
$v=41.8$ m/s
$a=4.66$ m/s$^2$
Work Step by Step
First, we solve for the distance as follows:
$x=2t^2$
$y=0.04t^3$
At $t=10$s, $x=200$m, $y=40$m
$d=\sqrt{200^2+40^2}=204$ m
$v_x=\frac{dx}{dt}=4t$
$a_x=\frac{dv_x}{dt}=4$
$v_y=\frac{dy}{dt}=0.12t^2$
$a_y=\frac{dv_y}{dt}=0.24t$
Now, find the velocity and acceleration at $t=10$ s,
$v=\sqrt{40^2+12^2}=41.8$ m/s
$a=\sqrt{4^2+2.4^2}=4.66$ m/s$^2$
