For an object whose velocity in ft/sec is given by v(t) = −2t^2 + 4, what is its distance travelled, in feet, on the interval t = 0 to t = 2 secs?

I think the answer is 2.667 but I am not one hundred percent sure.

the distance is

s = ∫[0,2] -2t^2 + 4 dt
= -2/3 t^3 + 4t [0,2]
= -16/3 + 8
= 8/3

you are correct

To find the distance traveled by the object on the interval t = 0 to t = 2 seconds, we need to integrate the velocity function over that interval.

The velocity function is given by v(t) = -2t^2 + 4.

To find the distance traveled, we need to integrate the absolute value of the velocity function.

∫[0 to 2] |v(t)| dt

Let's first find the absolute value of the velocity function:

|v(t)| = |-2t^2 + 4|

Since t is greater than or equal to zero, we can simplify the absolute value:

|v(t)| = -(-2t^2 + 4) = 2t^2 - 4

Now we can integrate the absolute value of the velocity function over the interval t = 0 to t = 2:

∫[0 to 2] (2t^2 - 4) dt

To integrate, we use the power rule:

∫(t^n) dt = (t^(n+1))/(n+1)

Applying the power rule, we get:

∫[0 to 2] (2t^2 - 4) dt = [(2/3)t^3 - 4t] [0 to 2]

Evaluating the definite integral, we get:

[(2/3)(2^3) - 4(2)] - [(2/3)(0^3) - 4(0)]
= [(16/3) - 8] - [0 - 0]
= (16/3) - 8
= (16/3) - (24/3)
= -8/3

Therefore, the distance travelled by the object on the interval t = 0 to t = 2 seconds is -8/3 feet.

To find the distance traveled by an object on a given interval, we need to integrate the velocity function over that interval. In this case, the velocity function is v(t) = -2t^2 + 4, and the interval is from t = 0 to t = 2 seconds.

To determine the distance traveled, we integrate the absolute value of the velocity function over the given interval. This is because the distance is always positive, regardless of the direction of motion.

∫|v(t)| dt = ∫| -2t^2 + 4| dt

To evaluate this integral, we need to break it down into two separate integrals based on the behavior of the velocity function:

1. For t values where -2t^2 + 4 > 0 (above the x-axis), the integral becomes:

∫(-2t^2 + 4) dt = [-2/3 * t^3 + 4t]

2. For t values where -2t^2 + 4 < 0 (below the x-axis), we multiply the velocity function by -1 to make it positive before integrating:

∫(2t^2 - 4) dt = [2/3 * t^3 - 4t]

Now, we evaluate both integrals over the interval t = 0 to t = 2:

Distance = ∫(-2/3 * t^3 + 4t) dt - ∫(2/3 * t^3 - 4t) dt

By substituting the upper and lower limits of integration:

Distance = [-2/3 * (2)^3 + 4(2)] - [-2/3 * (0)^3 + 4(0)] - [2/3 * (2)^3 - 4(2)] + [2/3 * (0)^3 - 4(0)]

Simplifying this expression, we get:

Distance = [-16/3 + 8] - [0 - 0] - [16/3 - 8] + [0 - 0]
Distance = [8/3] - [16/3] + [0] + [0]
Distance = -8/3

Therefore, the correct answer is -8/3 ft.

It appears that the value you provided, 2.667, is incorrect. The distance traveled is actually -8/3 ft.