Hill Bomb
Consider a skateboarder is bombing down a hill in a straight line so that his distance (D) from a point on the pavement after time (t) is D = 4 t^5 + 3 t. Determine the rider’s acceleration when t = 3.
Hint
Velocity is a derivative of distance.
Hint 2
Acceleration is a derivative of velocity.
Velocity is a derivative of distance.
$$$v=\frac{dr}{dt}$$$
where
$$v$$
is the instantaneous velocity,
$$t$$
is time, and
$$r$$
is position.
The power rule for the first derivative:
$$$\frac{d}{dx}[x^n]=n\cdot x^{n-1}$$$
Thus,
$$$D=4t^5+3t$$$
$$$v=(5)4t^{5-1}+(1)3t^{1-1}=20t^4+3t^0=20t^4+3$$$
Acceleration is a derivative of velocity.
$$$a=\frac{dv}{dt}$$$
where
$$a$$
is instantaneous acceleration,
$$t$$
is time, and
$$v$$
is instantaneous velocity.
Applying the power rule for the first derivative to the velocity function:
$$$a=(4)20t^{4-1}+0=80t^3$$$
Finally, substitute
$$t=3$$
:
$$$a=80(3)^3=80(27)=2,160$$$
2,160