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 = 3 secs?
9.000 <- my answer
−9.000
10.596
−3.00
good job
To find the displacement of the object on the interval t = 0 to t = 3 seconds, you need to calculate the definite integral of the velocity function v(t) = −t^2 + 6 over that interval.
The displacement is given by the integral of the velocity function:
Displacement = ∫[0 to 3] (-t^2 + 6) dt
To evaluate this integral, you can split it into two separate integrals:
Displacement = ∫[0 to 3] -t^2 dt + ∫[0 to 3] 6 dt
Integrating each term separately:
Displacement = [- (t^3)/3] evaluated from 0 to 3 + [6t] evaluated from 0 to 3
After plugging in the limits and simplifying, the displacement is:
Displacement = (- (3^3)/3 - 0) + (6(3) - 6(0))
Displacement = (- 27/3) + (18 - 0)
Displacement = (-9) + 18
Displacement = 9 feet
So, the correct answer is 9.000.
To find the displacement of the object on the interval from t = 0 to t = 3 seconds, we first need to find the antiderivative (also known as the indefinite integral) of the velocity function v(t).
The antiderivative of v(t) = -t^2 + 6 is given by the integral: S(v(t))dt.
Let's find the antiderivative of -t^2 + 6:
∫ (-t^2 + 6) dt = -1/3 * t^3 + 6t + C,
where C is the constant of integration.
Next, we evaluate the expression at the upper and lower limits, t = 0 and t = 3 seconds, respectively:
S(t=3) - S(t=0) = [-1/3 * (3)^3 + 6(3)] - [-1/3 * (0)^3 + 6(0)]
= [-1/3 * 27 + 18] - [0 + 0]
= [-9 + 18] - [0]
= 9 - 0
= 9 feet.
Therefore, the displacement of the object on the interval from t = 0 to t = 3 seconds is 9 feet.
So the correct answer is 9.000.