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

Question

Question options:

1)	7.67
2)	-3.33
3)	-5.33
4)	3.33

Answer 1

distance is the integral of velocity, so:

v(t) = -t^2 + 4

d(t) = -t^3 /3 + 4t + C

Evaluate for t = 0 to 4:

-4^3 /3 + 4*4 + C - 0 - 0 - C

-64/3 + 16

-16/3