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 = 0.5t3 + t2 + 2t, where x and t are expressed in feet and seconds, respectively. Determine the position, velocity, and acceleration of the boarder expressed when t = 12 seconds. (Round the final answer to one decimal place.)

To solve this question we need to know that the velocity v and acceleration a are the first and second derivatives of the position x respect to the time t.

Let's calculate these derivatives:

[tex]v=\frac{dx}{dt}=0.5\frac{d}{dt}(t^{3} )+\frac{d}{dt}(t^{2})+\frac{d}{dt}(2t)[/tex]

If we use the following formula:

[tex]\frac{d}{dt}(t^{n} )=n t^{n-1}[/tex] --- (1)

we get that:

[tex]v = 3*(0.5)t^{2} + 2t+2[/tex] --- (2)

Now the acceleration:

[tex]a=\frac{d^{2}x}{dt^{2}}=\frac{dv}{dt}=1.5\frac{d}{dt}(t^{2} )+\frac{d}{dt}(2t})+\frac{d}{dt}(2)[/tex]

Therefore:

[tex]a=3t+2[/tex] --- (3)

To determine the position, velocity and acceleration at t=12 we must evaluate these functions at t=12:

[tex]x(12) = 0.5*12^{3}+12^{2}+2*12\\v(12) = 1.5*12^{2}+2*12+2\\a(12)= 3*12+2[/tex]

Evaluating these equations we obtain the required values:

[tex]x(12s)=1032 feet\\v(12s)=242 feet/s\\a(12s)=38 feet/s^{2}[/tex]

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