Questions: The rate of change in the height of a falling object is a function of time given by d h/d t = 25 - 9.8 t Where d h/d t is in meters per second and t in seconds. Which of the following correctly represents the change in height of this object on the time interval [0,2] ? ○ ∫ from 0 to 2 of (25-9.8 t) d t ∫ from 2 to 0 of (25-9.8 t) ∫ from 0 to 2 of (25-9.8 t) ∫ from 2 to 0 of (25-9.8 t) d t Compute the change in height on [0,2]. Δ h =

Solution Steps

The rate of change in height of the falling object is given by the function: \[ \frac{d h}{d t} = 25 - 9.8 t \]

To find the change in height over the interval \([0, 2]\), we need to integrate this function with respect to \(t\) from 0 to 2. The correct integral representation is: \[ \int_{0}^{2} (25 - 9.8 t) \, dt \]

Thus, the correct choice is: \[ \boxed{\int_{0}^{2} (25 - 9.8 t) \, dt} \]

To find the change in height, we need to evaluate the integral: \[ \int_{0}^{2} (25 - 9.8 t) \, dt \]

First, we find the antiderivative of the integrand: \[ \int (25 - 9.8 t) \, dt = 25t - \frac{9.8 t^2}{2} + C \]

Next, we evaluate the antiderivative at the bounds 0 and 2: \[ \left[ 25t - \frac{9.8 t^2}{2} \right]_{0}^{2} \]

Substitute the upper bound \(t = 2\): \[ 25(2) - \frac{9.8 (2)^2}{2} = 50 - \frac{39.2}{2} = 50 - 19.6 = 30.4 \]

Substitute the lower bound \(t = 0\): \[ 25(0) - \frac{9.8 (0)^2}{2} = 0 \]

The change in height is the difference between the values at the upper and lower bounds: \[ \Delta h = 30.4 - 0 = 30.4 \]

Final Answer