Average Velocity

Question

Problem: h(t) = -16t^2 + 96t + 100, what is the average velocity between t = 4 and t = 5. Find the equation of the secant line between (4, h(4) and (5, h(5).

What I did:

[-16(5)^2 + 96(5) + 100] - [-16(4) + 96(4) + 100] / (5 - 4) and I got 180 - 420 / 1 so, -240 ft/s.

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I don't know how to find the secant equation, and I'm not sure if what I did is correct.

Answer 1

h(t)= -16t^2 +96t +100

velocity = v(t) = h'(t) = -32t+96

velocity v(4) = -32(4)+96 = -128+96= -32 ft/sec

v(5) = -32(5)+96 = -160+96=-64

average =(-32-64)/2 = -96/2 =- 48 ft/sec= secant line's slope between t=4 and 5

at t=4, h=-16(4^2+96(4)+100 = -256 +384 +100 = 128+100 = 228

at t=5, h= -16(5^2)+ 96(5) +100 = -400+ 580 +100= 180

secant line is the line from point (4,228) to (5,90)

with slope =

(180-228)/(5-4) = - 48 = (y-180)/(x-5)

y-180 = -48(x-5) is the secant line's equation in point slope form

y= -48x + 240+180

y =-48x +420 in slope intercept form

it's a downward opening parabola

or replace x with t, and y with h for time versus height which is the same

if graphed on the usual x and y axes

Answer 2

h(t) = h(5) + (- 240)(t - 5); h(t) = 180 - 240t + 1200; h(t) = 1380 - 240t.

This is equation of secant line h = 1380 - 240t in coordinate system tOh)