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 $$ s(t)=\frac{1}{100} t^{3}, 0 \leq t \leq 10 $$ (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) Average velocity: 1 m/s. (b) Instantaneous velocity: \(v(t) = \frac{3}{100}t^2\). (c) Instantaneous velocity equals average velocity at \(t \approx 5.77\) seconds.

Step 1: Calculate Average Velocity

The average velocity is defined as the change in position divided by the change in time. Therefore, we need to calculate \( \frac{s(10) - s(0)}{10 - 0} \). First, find \(s(10)\):\[s(10) = \frac{1}{100} \times 10^3 = 10\]Then, find \(s(0)\):\[s(0) = \frac{1}{100} \times 0^3 = 0\]Now, calculate the average velocity:\[\text{Average velocity} = \frac{10 - 0}{10 - 0} = 1 \text{ m/s}\]

Step 2: Determine Instantaneous Velocity Formula

The instantaneous velocity is given by the derivative of the position function \(s(t)\). To find the derivative \(s'(t)\), use the power rule:\[s(t) = \frac{1}{100} t^3 \]\[s'(t) = \frac{d}{dt}\left(\frac{1}{100} t^3\right) = \frac{1}{100} \times 3t^2 = \frac{3}{100} t^2\]Thus, the instantaneous velocity is \(v(t) = \frac{3}{100}t^2\).

Step 3: Find Time When Instantaneous Velocity Equals Average Velocity

Set the expression for instantaneous velocity equal to the average velocity to find \(t\):\[\frac{3}{100}t^2 = 1\]Solving for \(t^2\):\[3t^2 = 100 \]\[t^2 = \frac{100}{3}\]\[t = \sqrt{\frac{100}{3}}\]This simplifies to approximately \(t = 5.77 \text{ seconds}\).

Key Concepts

Understanding Average Velocity

When you think about average velocity, imagine it as the speed at which you would have to travel consistently to cover a certain distance in a certain amount of time.
It's crucial to note that it doesn't account for any variations in speed during the trip.

• Average velocity focuses on the overall change in position.
• It's calculated over a time interval, from a starting point, say time 0, to an ending point, say time 10 in this example.

In equation form, average velocity is given by:\[v_{\text{avg}} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}\]For our car problem, the average velocity from \(t = 0\) to \(t = 10\) seconds was straightforward: the car traveled a net 10 meters over 10 seconds, which simplifies to 1 m/s.
This gives you a broad view of the car's movement over that stretch of time.

Grasping Instantaneous Velocity

Instantaneous velocity, on the other hand, is like a snapshot of how fast something is moving at a specific moment.
Think about it like the speed you read on a speedometer at any given point.

• For instantaneous velocity, the shorter the time interval considered, the more accurate the measure becomes for that moment.
• Mathematically, it's captured by the derivative of the position function \(s(t)\) with respect to time \(t\).

The derivative tells us the rate of change of the position at any particular time \(t\), giving:\[v(t) = s'(t) = \frac{3}{100}t^2\]In our exercise, this allows us to calculate the instantaneous speed of the car at any moment between \(t = 0\) and \(t = 10\).
So, unlike average velocity, this value changes with time, responding to the specifics of the car's path, including any accelerations or decelerations.

Exploring Derivatives and Velocity

Derivatives in calculus play a vital role in describing how things change.
In our context of motion, they help us transition from a static description (position) to a dynamic description (velocity and, further, acceleration).

• The derivative \(s'(t)\) can be interpreted as the slope of the tangent line at any point on the position-time graph.
• This slope equates to the object’s instantaneous speed since it reflects how rapidly the position changes at a specific moment.

Integrating these concepts, if we take the derivative of our car’s position function \(s(t) = \frac{1}{100}t^3\), we compute:\[v(t) = s'(t) = \frac{3}{100}t^2\]This derivative gives us direct insight into the instantaneous behavior of the car.
It's the heart of understanding changing velocities, attachments of accelerations, and deeper changes over time in physics and beyond.
We used these derivatives to equate the instantaneous velocity to the average velocity, discovering that the car moved with the average speed at precisely 5.77 seconds. What a profound linkage between motion and mathematics!