## 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