Problem 9 - Calculus

Question

In a time of \(t\) seconds, a particle moves a distance of \(s\) meters from its starting point, where \(s=3 t^{2}\) (a) Find the average velocity between \(t=1\) and \(t=\) \(1+h\) if: (i) \(\quad h=0.1\) (ii) \(h=0.01,\) (iii) \(h=0.001\) (b) Use your answers to part (a) to estimate the instantaneous velocity of the particle at time \(t=1\)

Step-by-Step Solution

Answer

(a) Average velocities: 6 m/s for all \( h \) values; (b) Instantaneous velocity at \( t=1 \) is 6 m/s.

Step 1: Understand the Formula

The formula given is for the distance covered by the particle as a function of time: \( s = 3t^2 \). We need to find the velocity, which is the rate of change of distance with respect to time. For this exercise, we will calculate the average velocity over small intervals starting at \( t = 1 \).

Step 2: Average Velocity Formula

The average velocity \( v_{avg} \) over a time interval from \( t \) to \( t + h \) can be found using the formula \( v_{avg} = \frac{s(t+h) - s(t)}{h} \).

Step 3: Calculate \( s(1) \)

Find the position at \( t = 1 \). \( s(1) = 3(1)^2 = 3 \) meters.

Step 4: Calculate \( s(1+h) \)

Find the position at \( t = 1 + h \). This is \( s(1+h) = 3(1+h)^2 \).

Step 5: Expand \( s(1+h) = 3(1+h)^2 \)

Expand the expression: \( s(1+h) = 3(1^2 + 2h + h^2) = 3 + 6h + 3h^2 \) meters.

Step 6: Calculate Average Velocity for \( h = 0.1 \)

Substitute into the average velocity formula:\(v_{avg} = \frac{(3 + 6(0.1) + 3(0.1)^2) - 3}{0.1} = \frac{3.6 - 3}{0.1} = 6 \) meters per second.

Step 7: Calculate Average Velocity for \( h = 0.01 \)

Using the same method:\(v_{avg} = \frac{(3 + 6(0.01) + 3(0.01)^2) - 3}{0.01} = \frac{3.06 - 3}{0.01} = 6 \) meters per second.

Step 8: Calculate Average Velocity for \( h = 0.001 \)

Again, using the formula:\(v_{avg} = \frac{(3 + 6(0.001) + 3(0.001)^2) - 3}{0.001} = \frac{3.006 - 3}{0.001} = 6 \) meters per second.

Step 9: Determine Instantaneous Velocity

The instantaneous velocity at \( t=1 \) is the limit of the average velocity as \( h \) approaches \( 0 \), which from our previous calculations, appears to be consistently \( 6 \) meters per second, confirming the instantaneous velocity is \( 6 \) meters per second.

Key Concepts

Average Velocity

In calculus, understanding average velocity is crucial when analyzing the motion of an object over a period of time. Average velocity is the total distance traveled divided by the total time taken. It gives us a general sense of how fast an object is moving over a particular interval.

For our specific problem, we're looking at the motion of a particle described by the equation \( s = 3t^2 \). To find the average velocity between \( t=1 \) and \( t=1+h \), use the formula:

• \( v_{avg} = \frac{s(t+h) - s(t)}{h} \)

This formula calculates the change in position \( (s(t+h) - s(t)) \) over the change in time \( h \). This gives us the mean rate at which the particle is moving during this time interval.

Rate of Change

The concept of rate of change is fundamental in understanding how quantities vary over time. In the context of motion, the rate of change of position with respect to time is velocity.

For any function, the average rate of change between two points is calculated similarly to the average velocity. It measures how one quantity changes in relation to another over a specified interval, which in this case is how the distance \( s \) changes over time \( t \).

Since the function for our particle's motion is \( s = 3t^2 \), calculating the rate of change involves applying the average velocity formula over smaller and smaller intervals, ultimately leading us to think about instantaneous change.

Derivative

The derivative is a powerful mathematical tool that gives us the rate of change of a function at any given point. In motion problems, the derivative of the position function \( s(t) \) with respect to time \( t \) gives us the velocity function \( v(t) \).

For the function \( s = 3t^2 \), finding the derivative involves applying basic differentiation rules. The derivative \( \frac{ds}{dt} = 6t \) represents the velocity function. This tells us that the velocity of the particle at any given time \( t \) is \( 6t \) meters per second.

• The derivative opens the door to finding instantaneous rates of change effectively.
• It simplifies the process by providing an exact expression for the rate of change at any point rather than approximating over an interval.

Motion in Calculus

Motion in calculus involves studying how objects move and change positions over time, integrating multiple concepts we've discussed, like derivatives.

The precise description of motion requires understanding the relationship between position, velocity, and acceleration:

• Position is given by the function \( s(t) \).
• Velocity, the first derivative of position, is \( v(t) = \frac{ds}{dt} \).
• Acceleration, the derivative of velocity, gives \( a(t) = \frac{d^2s}{dt^2} \).

In our problem, finding the instantaneous velocity at \( t=1 \) involved taking the limit of the average velocity as \( h \) approaches zero, confirming that the motion of the particle is aligned with its velocity function. This seamless connection between concepts allows for a comprehensive understanding of how particles behave under various forces and conditions.