Consider the velocity function \( v(t)=2 \sin t \) for \( 0 \leq t \leq \pi \). Complete the sentence: The same distance could have been traveled over the given time period at a constant velocity of The same distance could have been traveled over the given time period at a constant velocity of \( \square \). (Type an exact answer in terms of \( \pi \).)

To find the constant velocity that would cover the same distance as the velocity function \( v(t) = 2 \sin t \) over the interval from \( 0 \) to \( \pi \), we first need to calculate the total distance traveled. The distance \( d \) is given by the integral of the velocity function from \( 0 \) to \( \pi \): \[ d = \int_{0}^{\pi} v(t) \, dt = \int_{0}^{\pi} 2 \sin t \, dt. \] Calculating this integral: \[ d = 2 \int_{0}^{\pi} \sin t \, dt. \] The integral of \( \sin t \) is \( -\cos t \), so we evaluate: \[ \int_{0}^{\pi} \sin t \, dt = [-\cos t]_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2. \] Thus, the distance is: \[ d = 2 \cdot 2 = 4. \] Now, to find the constant velocity that would cover this distance over the time interval from \( 0 \) to \( \pi \) (which has a duration of \( \pi \)), we use the formula for constant velocity: \[ v_{\text{constant}} = \frac{\text{distance}}{\text{time}} = \frac{4}{\pi}. \] Therefore, the same distance could have been traveled over the given time period at a constant velocity of \( \frac{4}{\pi} \).