Problem 47 - Calculus for Biology and Medicine

Question

A car moves in a straight line. At time \(t\) (measured in seconds), its position (measured in meters) is $$ s(t)=\frac{1}{10} t^{2}, 0 \leq t \leq 10 $$ (a) Find its average velocity between \(t=0\) and \(t=10\). (b) Find its instantaneous velocity for \(t \in(0,10)\). (c) At what time is the instantaneous velocity of the car equal to its average velocity?

Step-by-Step Solution

Answer

(a) 1 m/s; (b) \( \frac{1}{5}t \); (c) at \( t = 5 \) seconds.

Step 1: Finding Average Velocity

To find the average velocity between two points in time, we use the formula: \[ v_{\text{avg}} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \]For this problem, \(t_1 = 0\) and \(t_2 = 10\). First, calculate the positions at these times using the function \(s(t) = \frac{1}{10}t^2\).\[ s(0) = \frac{1}{10} \times 0^2 = 0 \]\[ s(10) = \frac{1}{10} \times 10^2 = 10 \]Now, substitute the position values into the average velocity formula:\[ v_{\text{avg}} = \frac{10 - 0}{10 - 0} = 1 \text{ m/s} \]

Step 2: Calculating Instantaneous Velocity

Instantaneous velocity is found by taking the derivative of the position function \(s(t)\) with respect to time \(t\). The derivative of \(s(t) = \frac{1}{10}t^2\) is:\[ v(t) = \frac{d}{dt} \left( \frac{1}{10}t^2 \right) = \frac{2}{10}t = \frac{1}{5}t \]Thus, the instantaneous velocity at time \(t\) is \(v(t) = \frac{1}{5}t\).

Step 3: Equating Instantaneous and Average Velocity

To find the time \(t\) when the instantaneous velocity equals the average velocity, set the expression for instantaneous velocity equal to the average velocity:\[ \frac{1}{5}t = 1 \]Solving for \(t\):\[ t = 5 \]This means that at \(t=5\) seconds, the instantaneous velocity and average velocity are equal.

Key Concepts

Average Velocity

The concept of average velocity is crucial in understanding motion over an interval of time. Average velocity represents the overall change in position divided by the time it takes for that change to occur. In mathematical terms, it is given by the formula:

• \( v_{\text{avg}} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \)

In this scenario, we are calculating the average velocity of a car over a 10-second interval, from \( t = 0 \) seconds to \( t = 10 \) seconds. The position function \( s(t) = \frac{1}{10} t^2 \) helps us determine the positions at these times. By plugging in the values, we find that the position at \( t = 0 \) is 0 meters, and at \( t = 10 \) it is 10 meters.
These calculations lead us to an average velocity of 1 m/s. This simple calculation gives us a sense of how fast the car is moving, on average, during that entire period.

Instantaneous Velocity

Where average velocity provides an overall view of speed over a duration, instantaneous velocity gives us the speed of an object at a specific moment in time. This concept is incredibly important in calculus and is determined using derivatives. For the car's position function \( s(t) = \frac{1}{10} t^2 \), we derive with respect to time \( t \) to find its instantaneous velocity.

• The derivative \( v(t) = \frac{d}{dt}\left( \frac{1}{10} t^2 \right) = \frac{2}{10} t = \frac{1}{5} t \) represents the instantaneous velocity function.

This means at any given time \( t \), the car's speed can be directly calculated by the expression \( v(t) = \frac{1}{5} t \), showing how the speed of the car changes at every second between \( t = 0 \) and \( t = 10 \).

Derivative

Derivatives are at the heart of calculus. They provide the rate at which one quantity changes with respect to another. When we talk about velocity in the context of calculus, the rate of change of the position function with respect to time is the derivative.

• For a given function \( s(t) = \frac{1}{10} t^2 \), the derivative \( v(t) = \frac{1}{5} t \) tells us the instantaneous velocity.

Derivatives reveal how a function grows or decreases at any point. In other contexts, it represents things like acceleration, slopes of curves, and more. Thus, the derivative can be visualized as a slope on a graph. It's a powerful concept that extends well past simple velocity calculations and is crucial in real-world applications.

Position Function

The position function is a fundamental representation of motion and is used to specify the location of an object at any time \( t \). In our example, the position function given is \( s(t) = \frac{1}{10} t^2 \).

• This function tells us how the car's position changes as time progresses, providing a continuous journey description plotted over time.

Understanding the position function is key to studying motion effectively. It allows us to calculate other essential components of motion such as velocity and acceleration, by using calculus tools like derivatives and integrals. This specific function shows a simple yet profound example of a quadratic relationship, common in constant acceleration scenarios like our car's movement.