3:For an object whose velocity in ft/sec is given by v(t) = −2t^2 + 4, what is its distance travelled, in feet, on the interval t = 0 to t = 2 secs?

Question

a:4.876 b:2.667 c:9.752 d:2.438

Answer 1

then the distance formula is

s = (-2/3)t^3 + 4t + c, where c is a constant

at t = 0, s = c
at t = 2, s = -16/3 + 8 + c

distance in the given interval
= -16/3 + 8 + c - c
= 8/3 which is close to choice b)

Answer 2

^^^thats wrong

Answer 3

the answer is 4.876
I took the exam and put 2.667 and got it wrong so here's the right answer