For an object whose velocity in ft/sec is given by v(t) = -2t^2 + 8, what is its distance travelled, in feet, on the interval t = 0 to t = 3 secs?

Question

Question options:

1)	6.00
2)	0.67
3)	15.33
4)	5.11

Answer 1

V(t) = -2t² +8

∫v(t) = ∫(-2t² + 8 )dt = d

d= [ -(2/3)t³ + 8t ] t in interval 0 to 3

d= [ - (2/3)(3³) + 8(3) ] - [ - (2/3)(0³) + 8(0) ]

d = [ - (0.6667)(27) + 24 ] - 0

d= [ -18 + 24 ]

d = 6

The distance is 6 ft

Answer 2

AT T=0, CALCULATE V, V0=-2(0)+8=8 FEET/SEC

AT T=3 SEC,V3= -2(3)^2+8 = -10

DISTANCE= |8-10|/(3-0)=2/3=0.67 FEET