For an object whose velocity in ft/sec is given by v(t) = −3t2 + 6, what is its distance travelled, in feet, on the interval t = 0 to t = 2 secs?
To find the distance traveled by an object on the interval from t = 0 to t = 2 seconds, we need to find the definite integral of the absolute value of the velocity function v(t) from t = 0 to t = 2.
First, let's write the absolute value of v(t):
|v(t)| = |-3t^2 + 6|
Now, let's consider the two cases when v(t) is positive and negative separately.
When v(t) > 0 (t < sqrt(2)), the absolute value can be removed:
|v(t)| = -(-3t^2 + 6) = 3t^2 - 6
When v(t) < 0 (t > sqrt(2)), the absolute value becomes negative:
|v(t)| = -(3t^2 - 6) = -3t^2 + 6
Therefore, the distance traveled by the object on the interval t = 0 to t = 2 seconds is given by the integral of the absolute value of v(t):
Distance = ∫ |v(t)| dt from t = 0 to t = 2
= ∫ (3t^2 - 6) dt from t = 0 to t = 2 (for v(t) > 0)
+ ∫ (-3t^2 + 6) dt from t = 2 to t = 2 (for v(t) < 0)
Evaluating the integral in parts:
Distance = [t^3 - 6t] from t = 0 to t = 2 (for v(t) > 0)
+ [-t^3 + 6t] from t = 2 to t = 2 (for v(t) < 0)
Now, substitute the upper and lower limits of integration:
Distance = [(2)^3 - 6(2)] - [(0)^3 - 6(0)] (for v(t) > 0)
+ [-(2)^3 + 6(2)] - [-(2)^3 + 6(2)] (for v(t) < 0)
Simplifying the equation further:
Distance = [8 - 12] - [0 - 0] (for v(t) > 0)
+ [-8 + 12] - [-8 + 12] (for v(t) < 0)
Distance = -4 (for v(t) > 0)
+ 4 (for v(t) < 0)
Adding the distances for both cases:
Distance = -4 + 4
= 0
Therefore, the object traveled a distance of 0 feet on the interval t = 0 to t = 2 seconds.
To find the distance traveled by an object on the interval t = 0 to t = 2 seconds, we need to integrate the velocity function over this interval.
The distance traveled is obtained by integrating the absolute value of the velocity function. In this case, the absolute value of v(t) is given by |v(t)| = |-3t^2 + 6|.
Now, let's break down the problem into two intervals: t = 0 to t = 1, and t = 1 to t = 2.
1) For the interval t = 0 to t = 1:
On this interval, the velocity function v(t) = -3t^2 + 6 is negative because -3t^2 is always negative. Hence, we can simplify the integral to ∫(-v(t))dt = ∫(3t^2 - 6)dt. Integrating this function from t = 0 to t = 1 gives us the negative of the area under the curve.
∫(3t^2 - 6)dt = t^3 - 6t ∣[0, 1] = (1^3 - 6(1)) - (0^3 - 6(0)) = 1 - 6 = -5 ft.
2) For the interval t = 1 to t = 2:
On this interval, the velocity function v(t) = -3t^2 + 6 is positive because -3t^2 is always positive. We can directly integrate the function from t = 1 to t = 2.
∫(3t^2 - 6)dt = t^3 - 6t ∣[1, 2] = (2^3 - 6(2)) - (1^3 - 6(1)) = 8 - 12 - 1 + 6 = 1 ft.
To find the total distance traveled, we add the absolute values of the two results:
|-5| + 1 = 5 + 1 = 6 ft.
Therefore, the object travels a total distance of 6 feet on the interval t = 0 to t = 2 seconds.