The distance d in metrers that a ball has dropped at t seconds follows the equation f(t) = 4.9^2. If the ball is dropped from a height of 54 m, find the rate of change of distance with respect to time at 3 s.
2007-04-04 04:46:20 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Your formula should be
f(t)= -4.9t^2+54
de/dt=-9.8t =-29.4m/s at t=3s
The - sign means that the velocity is directed downwards
2007-04-04 04:57:00 · answer #1 · answered by santmann2002 7 · 0⤊ 0⤋
Sounds like a trick question (meaning there's more information there than you need to answer the question).
Assuming the initial velocity of the ball is zero, the velocity ("rate of change of distance") after 3 seconds would follow the equation v=at (velocity equals acceleration x time.
Acceleration due to gravity is 9.8m/s/s. So the velocity after 3 seconds would be 9.8 x 3 (m/s) = 29.4m/s.
This ofcourse assumes that we're talking about our planet's gravity, with no air resistance involved.
2007-04-04 11:59:56 · answer #2 · answered by Daniel M 1 · 0⤊ 0⤋
27
2007-04-04 11:53:37 · answer #3 · answered by joe 1 · 0⤊ 0⤋
you should've labeled this 'attention all asians' then you'd get some results
2007-04-04 11:53:39 · answer #4 · answered by diddy kong 1 · 0⤊ 0⤋