# A snowboarder starts from rest at the top

A snowboarder starts from rest at the top of a double black diamond hill. As he rides down the slope, GPS coordinates are used to determine his displacement as a function of time: $x= 0.5t^3 + t^2 + 2t$ where x and t are expressed in ft and seconds, respectively. Determine the position, velocity, and acceleration of the boarder when $t=5$ seconds.

#### Solution:

We can figure out the position by substituting 5 seconds into our position equation.

$x= 0.5t^3 + t^2 + 2t$

$x= 0.5(5)^3 + (5)^2 + 2(5)=97.5$ ft

To figure out the velocity, we will take the derivative of our position equation.

$v=\dfrac{\text{d}x}{\text{d}t}=1.5t^2+2t+2$

Again, substitute $t=5$ into our velocity equation to figure out the velocity at 5 seconds.

$v=1.5t^2+2t+2$

$v=1.5(5)^2+2(5)+2=49.5$ ft/s

To find the acceleration, we will take the derivative of the velocity equation.

$a=\dfrac{\text{d}v}{\text{d}t}=3t+2$

As before, substitute $t=5$ into our acceleration equation to figure out the acceleration at 5 seconds.

$a=3t+2$

$a=3(5)+2=17\dfrac{\text{ft}}{s^2}$

$x=97.5$ ft
$v=49.5$ ft/s
$a=17\dfrac{\text{ft}}{s^2}$