(b) Draw the secant line of s(t) from t = 1 to t = 1.5. Without doing any additional work, what is the slope of that line?

Question

position function s(t) = −16t^2 + 200t

Answer 1

same as average of the tangent lines at t=1 and t=1.5

= derivative of s(t) = s'(t) = -32t +200

= average of

-32+200 = 168

and -32(1.5)+200 = -48+200 = 152

= (168+152)/2 = 84+76 = 160

which is far less "work" than calculating the secant line slope as (s(1.5) -s(1))/(1.5-1) which leads to the identical slope