Question

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?

Answer 1

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

\[\sqrt{18.75}= t\]