Question

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 = 3 secs?
7.67
−3.00
−0.33
3.00

Answer 1

Answer:

The answer is 3 feet

Step-by-step explanation:

The displacement  function is the integration of the velocity formula, so:

$d(t)=\int\limits^0_t \, (-t^{2} +4) dt \\d(t)=\int\limits^0_t{} \, (-t^{2}) dt + \int\limits^0_t{} \, 4 dt \\\\d(t)=(\frac{-t^{3} }{3} + C ) + (4t + C)\\\\\\\ d(t)=frac{-t^{3} }{3} + 4t + C$

if we evaluate the displacement for the upper and lower limit it would be

D = total displacement = d(3) - d(0)

$D=(\frac{-3^{3} }{3} + 12 + C) - (\frac{-0^{3} }{3} + 4*0 + C)\\\\D=(-9 + 12 + C) - (0 + 0 + C)\\\\\\D=(3 + C) - C\\\\\\D=3$

Answer 2

Its either A or D if its wrong sorry hope this helped tho