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  • 5 years ago

If I drop a car from a height of 300 feet straight down. It's velocity after t seconds is -32t feet/second. How do I determine the position function? How fast is the car going dropping after four seconds? How far has the car dropped after 4 seconds? HOw many seconds will it take for the car to splat on the ground?

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  1. anonymous
    • 5 years ago
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    To get the position function take the anti derivative of -32t dx/dt = -32t x= -16 t^2 + constant to find that constant you need some given information and what you can assume is that in t=0 you have 0 seconds but then your not moving at all so you can say at t=0 your position (x) is 300 so plug that in to the position function to find the constant 300= 0 + constant so you know the constant is 300 the function for position is x= - 16t ^2 + 300 Ok for the second part you have t-4 but you want to find x so plug that in to your position function and you get x= -16 4^2 + 300 x= (-16) * (16) + 300 x = -256 + 300 x= 44 so you are 44 feet off the ground after 4 seconds for the last part you want to know when x=0 or when you hit the ground but you want to find t 0= -16 t^2 + 300 -300= -16 t^2 -300/ -16= t^2 18.75= t^2 so the final answer is \[\sqrt{18.75}= t\]

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