Asked by matt
(Compute all instantaneous rate of change using the limit of the difference quotient)
2) A football's path is represented by the function h(t)=-4.9t^2 +10t+2, where h is its height, in meters, after t seconds. Find the rate of change of the football's height at 1s.
2) A football's path is represented by the function h(t)=-4.9t^2 +10t+2, where h is its height, in meters, after t seconds. Find the rate of change of the football's height at 1s.
Answers
Answered by
bobpursley
lim (-4.9(t+dt)^2+10(t+dt)+2)/dt
= -4.9(t^2+2tdt+dt^2)+10t+10dt+2 +4.9t^2-10t-2) /dt
= -4.9 (0t^2 + 0t+2tdt+dt^2+10dt) /dt
= -9.8t +10 as the limit of dt>>>zero
at t=1, then rate of change=.2 m/s
= -4.9(t^2+2tdt+dt^2)+10t+10dt+2 +4.9t^2-10t-2) /dt
= -4.9 (0t^2 + 0t+2tdt+dt^2+10dt) /dt
= -9.8t +10 as the limit of dt>>>zero
at t=1, then rate of change=.2 m/s
Answered by
Damon
dh/dt = -9.8 t + 10
when t = 1
dh/dt = 10-9.8 = 0.2
when t = 1
dh/dt = 10-9.8 = 0.2
Answered by
matt
thats 0.2m/s right?
Answered by
matt
ok got it thanks
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