For an object whose velocity in ft/sec is given by v(t) = −3t2 + 5, what is its displacement, in feet, on the interval t = 0 to t = 2 secs?

Question

For an object whose velocity in ft/sec is given by v(t) = −3t² + 5, what is its displacement, in feet, on the interval t = 0 to t = 2 secs?

Answer 1

v= -3t^2 +5

s(t) = integral of v(t) = -3t^3/3 +5t = -t^3 +5t + c

evalutated at 0 to 2

= -(2)^3 +5(2) = -8 +10

= 2 feet

but v(t) = -3t^2 +5 = 0 when

t^2= 5/3= 1 2/3

from t = 0 to t = 1 2/3 it moves forward

then from 1 2/3 to 2 it moves backward

at t =5/3 s= -3t^2+5 = -(5/3)^2 +5 = -25/9 +5= 45/9- 25/9= 40/9 = 4 4/9

at t = 2, s= 2

net displacement = 4 4/9 -2 = 2 4/9

about 2.44 feet

4.44 steps forward, 2 backwards, net 2.44 forward displacment