Asked by Brett
A test tube is knocked off a tower at the top of a building that is 400 feet above the ground. (For our purposes, we will assume that air resistance is negligible.) The test tube drops 16𝑡^2 feet in 𝑡 seconds.
a) Calculate the average velocity in the last two seconds of the fall.
b) Calculate the instantaneous velocity when the test tube lands.
a) Calculate the average velocity in the last two seconds of the fall.
b) Calculate the instantaneous velocity when the test tube lands.
Answers
Answered by
bobpursley
so first is to find time of fall:
s=400-16t^2
at final time, s is zero
or Tf=sqrt(400/16)=5 seconds.
velocity= ds/dt=-32*t
veloicty(5)=-160 m/s
velocity(3)=-96 m/s
so what is the average velocity?
s=400-16t^2
at final time, s is zero
or Tf=sqrt(400/16)=5 seconds.
velocity= ds/dt=-32*t
veloicty(5)=-160 m/s
velocity(3)=-96 m/s
so what is the average velocity?
Answered by
Brett
Thank you, for some reason I thought that 16t^2 was already acceleration (meaning f''(t) ) so I was like "how do I back-track to get velocity" ( f'(t) ).
Anyway, I worked out -128ft/s for V_avg (since the equation is asking for feet/second).
Thus the time hitting the ground was f'(5).
Thanks for the help Bob!
Anyway, I worked out -128ft/s for V_avg (since the equation is asking for feet/second).
Thus the time hitting the ground was f'(5).
Thanks for the help Bob!
Answered by
oobleck
I haven't looked through the calculations, but f'(5) is not time -- it is velocity at t=5.
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