For an object whose velocity in ft/sec is given by v(t) = -t2 + 6, what is its displacement, in feet, on the interval t = 0 to t = 6 secs?

Question

For an object whose velocity in ft/sec is given by v(t) = -t² + 6, what is its displacement, in feet, on the interval t = 0 to t = 6 secs?

A	-36
B	36
C	0
D	55.596

Answer 1

d(t) = integral of v(t) = -t^3/3 + 6t evaluated at 0 to 6 = -6^3/3 + 36 = 3=-72 + 36 = -36 feet