60–63. Equivalent constant velocity Consider the following velocity functions. In each case, complete the sentence: The same distance could have been traveled over the given time period at a constant velocity of ________.


v(t) = t(25−t²)^1/2, for 0≤t≤5

Verified step by step guidance

1. Identify the time interval over which the velocity function is defined, which is from \(t=0\) to \(t=5\) seconds.
2. Recall that the total distance traveled over the time interval can be found by integrating the velocity function \(v(t)\) with respect to time: \(\text{Distance} = \int_0^5 v(t) \, dt\).
3. Set up the integral for the given velocity function: \(\int_0^5 t \sqrt{25 - t^2} \, dt\).
4. Calculate the definite integral to find the total distance traveled during the time interval. (You can use substitution or other integration techniques to solve this integral.)
5. Once the total distance is found, find the equivalent constant velocity \(v_c\) by dividing the total distance by the total time interval: \(v_c = \frac{\text{Distance}}{5}\).

Key Concepts

Average Velocity

Average velocity over a time interval is defined as the total displacement divided by the total time. It represents a constant velocity that would cover the same distance in the same time period, making it essential for comparing variable velocity functions to a constant velocity.

Definite Integral for Displacement

The definite integral of a velocity function over a time interval gives the total displacement traveled during that period. Calculating this integral is necessary to find the total distance covered, which is then used to determine the equivalent constant velocity.

Velocity Function and Domain

Understanding the given velocity function v(t) = t(25−t²)^(1/2) and its domain 0 ≤ t ≤ 5 is crucial. This function describes how velocity changes over time, and recognizing its behavior helps in setting up the integral and interpreting the physical meaning of the problem.