Question

9.

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? (5 points)
-36
36
0
55.596

Answer 1

The displacement of the object with velocity function v(t) = -t^2 + 6 on the interval from t = 0 to t = 6 seconds is -36 feet.

Step-by-step explanation:

To determine the displacement of an object whose velocity function is v(t) = -t2 + 6, we need to integrate the velocity function over the given time interval. The displacement is given by the integral of the velocity function from t = 0 to t = 6 seconds.

The integral of v(t) is the position function s(t):

So the position function s(t), which gives the displacement when integrated from 0 to 6, is:

s(t) = -t3/3 + 6t

Evaluating s(t) at t = 6 and subtracting the value at t = 0 gives us the displacement:

Displacement = s(6) - s(0) = (-63/3) + 6(6) - (0) = -72 + 36 = -36 feet