The acceleration of the position function s(t) = 4t^2– 8t + 2 when t = 7.

Question

a) 7
b) 48
c) 142
d) 8

Answer 1

s(t) = 4t^2– 8t + 2 when t = 7
= 4(7)^2-8(7)+2
= 4(49)-56+2
= 196-54

Answer 2

s(t) = 4t^2– 8t + 2 when t = 7
= 4(7)^2-8(7)+2
= 4(49)-56+2
= 196-54
= 140+2
= 142

Answer 3

acceleration is the 2nd derivative of s(t)

s(t) = 4t^2– 8t + 2
v(t) = 8t - 8
a(t) = 8

since the acceleration is constant,
when t = 7, a = 8 units of distance/(units of time)^2