let s(t)= t^4-4t^3+8t for 0 ≤ t ≤ 4
find the total distance traveled by the object.
find where the acceleration is positive.
To find the total distance traveled by the object, we need to compute the definite integral of the absolute value of the velocity function over the given interval.
First, let's find the velocity function, v(t), by taking the derivative of the position function, s(t).
v(t) = s'(t) = 4t^3 - 12t^2 + 8
To find where the acceleration is positive, we need to determine the intervals where the second derivative of the position function, s(t), is positive.
Let's proceed with solving both parts step by step.
1. Total distance traveled:
To find the total distance traveled, we need to calculate the definite integral of the absolute value of the velocity function over the interval [0, 4].
∫[0,4] |v(t)| dt
To do this, we need to break down the interval [0, 4] into smaller intervals based on any points where the velocity function changes direction (crosses the x-axis).
Since the velocity function is a cubic polynomial, it is non-negative over the entire interval [0, 4], meaning it doesn't change direction. Therefore, we can integrate the velocity function directly over the interval [0, 4].
∫[0,4] |4t^3 - 12t^2 + 8| dt
To compute this integral, we need to evaluate the antiderivative (indefinite integral) of the absolute value of the velocity function and then subtract the values at the endpoints of the interval.
To find the antiderivative of the absolute value of the velocity function, we need to consider two cases:
Case 1: When the value inside the absolute value is non-negative (positive or zero).
∫ |4t^3 - 12t^2 + 8| dt = ∫ (4t^3 - 12t^2 + 8) dt = t^4 - 4t^3 + 8t + C
Case 2: When the value inside the absolute value is negative.
In this case, we need to multiply the value inside the absolute value by -1 to make it non-negative, and then integrate it.
∫ |4t^3 - 12t^2 + 8| dt = ∫ (-4t^3 + 12t^2 - 8) dt = -t^4 + 4t^3 - 8t + C'
Now, we can substitute the endpoints of the interval [0, 4] into the antiderivative and subtract the values.
Total distance traveled = (4^4 - 4(4^3) + 8(4)) - (0^4 - 4(0^3) + 8(0))
Simplifying: Total distance traveled = (256 - 4(64) + 32) - (0 - 0 + 0)
Total distance traveled = 256 - 256 + 32 - 0
Total distance traveled = 32 units
Therefore, the total distance traveled by the object is 32 units.
2. Where the acceleration is positive:
To find where the acceleration is positive, we need to determine the intervals where the second derivative of the position function, s(t), is positive.
Let's find the second derivative of the position function, s(t), which will give us the acceleration function, a(t).
a(t) = s''(t) = 12t^2 - 24t
Now, let's determine the intervals where the acceleration is positive.
To find these intervals, we need to solve the inequality a(t) > 0.
12t^2 - 24t > 0
We can factor out a common factor of 12t, which gives:
12t(t - 2) > 0
Now, we set each factor equal to zero and find the critical points:
12t = 0 => t = 0
t - 2 = 0 => t = 2
These critical points divide the number line into three intervals: (-∞, 0), (0, 2), and (2, ∞).
We can choose a test point in each interval and evaluate the sign of the expression 12t(t - 2) to determine the sign of the function in that interval.
For t < 0, let's choose t = -1 (a test point):
12(-1)(-1 - 2) = -12 < 0
For 0 < t < 2, let's choose t = 1 (a test point):
12(1)(1 - 2) = 12(-1) = -12 < 0
For t > 2, let's choose t = 3 (a test point):
12(3)(3 - 2) = 36 > 0
From the signs of the expressions, we can conclude that the acceleration is positive on the interval (2, ∞).
Therefore, the object has a positive acceleration when 2 < t ≤ 4.