## 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 riderâ€™s acceleration when t = 3.

##
__
__**Hint**

**Hint**

Velocity is a derivative of distance.

##
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__**Hint 2**

**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