Problem 48 - 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) Instantaneous velocity is \( \frac{1}{5}t \); (c) At \(t = 5\) s.

Step 1: Understand Average Velocity Formula

Average velocity is calculated as the change in position divided by the change in time. For a function \(s(t)\), the average velocity over a time interval \([a, b]\) is given by the formula: \[ v_{avg} = \frac{s(b) - s(a)}{b - a} \].

Step 2: Calculate s(t) at Boundary Points

Substitute the boundary points into the position function \( s(t) = \frac{1}{10}t^2 \): - At \( t = 0 \), \( s(0) = \frac{1}{10} \times 0^2 = 0 \). - At \( t = 10 \), \( s(10) = \frac{1}{10} \times 10^2 = 10 \).

Step 3: Compute Average Velocity

Using the calculated positions, apply the average velocity formula: \[ v_{avg} = \frac{s(10) - s(0)}{10 - 0} = \frac{10 - 0}{10} = 1 \text{ m/s} \].

Step 4: Understand Instantaneous Velocity as Derivative

Instantaneous velocity at a time \( t \) is given by the derivative of the position function \( s(t) \). For the function \( s(t) = \frac{1}{10}t^2 \), the derivative is \( s'(t) \).

Step 5: Compute Derivative of s(t)

Differentiate \( s(t) = \frac{1}{10}t^2 \) with respect to \( t \) to find \( s'(t) \): \[ s'(t) = \frac{d}{dt}\left(\frac{1}{10}t^2\right) = \frac{2}{10}t = \frac{1}{5}t \].

Step 6: Express Instantaneous Velocity

The instantaneous velocity \( v(t) \) for any \( t \in (0, 10) \) is given by \( v(t) = s'(t) = \frac{1}{5}t \).

Step 7: Find Time When Instantaneous Velocity Equals Average Velocity

Set the instantaneous velocity equal to the average velocity and solve for \( t \): \[ \frac{1}{5}t = 1 \] Solving gives \( t = 5 \). At \( t = 5 \) seconds, the instantaneous velocity equals the average velocity.

Key Concepts

Average Velocity

Average velocity is a fundamental concept in calculus for biology and medicine, as well as physics. It represents the overall change in position over a specific time interval.

To find average velocity, you calculate the difference in position between two points and divide this by the time elapsed between those points. This is often used to understand the trend of an object's movement, instead of its speed at a particular moment.

• The formula is: \[ v_{\text{avg}} = \frac{s(b) - s(a)}{b - a} \]
• It applies to the motion over an interval \([a, b]\).

In our exercise, the car's movement is described from time \(t=0\) to \(t=10\), and its position function is given by \(s(t) = \frac{1}{10}t^2\). By plugging the time boundaries into the position function, you find that:

• At \(t = 0\), the position \(s(0) = 0\) meters.
• At \(t = 10\), the position \(s(10) = 10\) meters.

So, the average velocity between these two times is \(1\text{ m/s}\). This tells us about the car's overall motion as it moved straight.

Instantaneous Velocity

Instantaneous velocity differs from average velocity as it refers to the speed and direction of an object at a specific moment in time. It is essentially the object's speedometer reading at that point.

To find the instantaneous velocity, you must determine the derivative of the position function. The derivative gives a "snapshot" of the rate of change of position with respect to time at any instant.
The position function for the car in our example is \(s(t) = \frac{1}{10}t^2\). By taking the derivative, we find:

• The derivative function is: \[ s'(t) = \frac{d}{dt}\left(\frac{1}{10}t^2\right) = \frac{2}{10}t = \frac{1}{5}t \]
• Substitute any \(t\) value within the range to get instantaneous velocity.

This derivative function, \(s'(t) = \frac{1}{5}t\), provides the instantaneous velocity of the car at any time \(t\). You can see how the velocity changes as time progresses.

Derivative of Position Function

The concept of the derivative is central to calculus for biology and medicine. It helps us understand changes in dynamic systems and processes over time.

In our context, the position function \(s(t) = \frac{1}{10}t^2\) gives the location of the car at any time \(t\). The derivative, noted as \(s'(t)\), indicates how quickly the position is changing at any time. It essentially tells us the instantaneous velocity at each moment.
To compute the derivative, apply the power rule for differentiation, which involves multiplying the power of time \(t\) by the coefficient and subtracting one from the power:

• Derivative of the function: \[ s'(t) = \frac{2}{10}t = \frac{1}{5}t \]
• This process finds the "rate of change" or slope of the line at any point on the function \(s(t)\).

In simpler terms, while the position function gives you the location, its derivative tells you about the speed at which this location is changing at any instant. The car's changing velocity is linked directly to the derivative of the position function.