Question

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 _____. $$v(t)=t\left(25-t^{2}\right)^{1 / 2}, \text { for } 0 \leq t \leq 5$$

Short answer

Answer: The constant velocity required is $$\frac{125}{15}$$ m/s.

Step-by-step solution

Integrate the velocity function

Find the integral of the velocity function with respect to time (t) between 0 and 5: $$\int_{0}^{5} v(t) dt = \int_{0}^{5} t\left(25-t^{2}\right)^{1/2} dt$$

Perform substitution

To make it easier to integrate, let's perform a substitution: Let $$u = 25 - t^2 \Rightarrow du = -2t dt$$ We also need to update the limits of the integration: When $$t = 0, u = 25$$ and when $$t = 5, u = 0$$ Now, our integral with respect to u: $$\frac{1}{2} \int_{25}^{0} u^{1/2} du$$

Integrate and evaluate the integral

Now we can integrate with respect to u: $$\frac{1}{2} \left[\frac{2}{3} u^{3/2}\right]_{25}^{0} = \frac{1}{3} \left[u^{3/2}\right]_{25}^{0}$$ Evaluate the integral: $$\frac{1}{3}\left(0 - 25^{3/2}\right) = -\frac{25\sqrt{25}}{3} = -\frac{125}{3}$$ Since the distance traveled must be a positive value, we take the absolute value of the result: $$distance = \frac{125}{3}$$

Calculate the constant velocity

To find the constant velocity needed to travel the same distance over the given time period, divide the distance by the total time (5 seconds): $$constant\,velocity = \frac{distance}{time} = \frac{\frac{125}{3}}{5} = \frac{125}{15}$$ The same distance could have been traveled over the given time period at a constant velocity of $$\frac{125}{15}$$.

Key concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus, often used to find areas under curves. In the context of velocity functions, they help calculate the total distance traveled over a given time period. When you integrate a velocity function with respect to time, you sum up all the infinitesimal distances covered over time, resulting in the total distance.
To compute the definite integral for a velocity function like \(v(t) = t(25-t^2)^{1/2}\), you need to evaluate it between specific time limits, say from \(t = 0\) to \(t = 5\). By doing so, you find the net distance traveled by the object. This process involves applying fundamental integration techniques either manually or using calculators to evaluate it after finding an antiderivative.
In essence, the integral of a velocity function gives the net change in position, which is the distance traveled during that interval.

Substitution Method

The substitution method is a powerful technique in calculus used to simplify the integration process. It involves changing the variable of integration to make the integral easier to solve. This is particularly useful for integrals involving complex expressions, like square roots or products.
Let’s see how substitution is applied: Using the velocity function \(v(t) = t(25-t^2)^{1/2}\), a substitution is made by letting \(u = 25 - t^2\). Calculating the derivative, we find \(du = -2t \, dt\). This helps transform the integral from a challenging form with respect to \(t\) into an easier one with respect to \(u\).
By updating the limits of integration, from \(t = 0\) translating to \(u = 25\) and \(t = 5\) to \(u = 0\), the integral becomes more straightforward to evaluate. This makes tough problems approachable by reducing them into simpler standard forms.

Average Velocity

Average velocity is a concept that helps us understand overall motion over a period of time. It is computed as the total distance traveled divided by the total time taken. In our example, once we’ve found the distance using the definite integral, calculating the average velocity gives us a constant speed.
This means if an object travels the same distance at this constant speed over the specific time, it results in the same travel characteristics as the variable velocity. The average velocity provides a simplified and more accessible way to describe motion, especially useful when planning travel or solving physics problems.
In practical terms, once the distance is found to be \(\frac{125}{3}\), dividing by the time \(5\) seconds, as in our example, yields the average speed. This conversion highlights how dynamic motions in nature can be depicted in steady, everyday terms.