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:

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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}

 

Final Answers:

x=97.5 ft

v=49.5 ft/s

a=17\dfrac{\text{ft}}{s^2}

 

This question can be found in Vector Mechanics for Engineers: Statics and Dynamics, 11th edition, chapter 11, question 11.1.

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