Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 32 ft/s from a height of 48 ft above the ground. The height (in feet) of the stone above the ground t seconds after it is thrown is s(t) = -16t² + 32t + 48.
On what intervals is the speed increasing?

Verified step by step guidance

1. Step 1: Understand that the speed of the stone is increasing when the magnitude of its velocity is increasing. The velocity function is the derivative of the height function s(t).
2. Step 2: Differentiate the height function s(t) = -16t^2 + 32t + 48 to find the velocity function v(t). The derivative is v(t) = s'(t) = -32t + 32.
3. Step 3: Determine when the velocity is increasing. This occurs when the acceleration, which is the derivative of the velocity function, is positive. Differentiate v(t) to find the acceleration function a(t).
4. Step 4: The acceleration function is a(t) = v'(t) = -32. Since this is a constant negative value, the velocity is not increasing at any point.
5. Step 5: Conclude that the speed of the stone is not increasing at any interval because the acceleration is constant and negative, indicating the velocity is decreasing.