Problem 4 - Active Calculus

Question

The displacement (in meters) of a particle moving in a straight line is given by $$s=t^{2}-5 t+12$$ where \(t\) is measured in seconds. (A) (i) Find the average velocity over the time interval [3,4] . Average Velocity \(=\square\) (ii) Find the average velocity over the time interval [3.5,4]. Average Velocity \(=\square\) (iii) Find the average velocity over the time interval [4,5] . Average Velocity \(=\square\) (iv) Find the average velocity over the time interval [4,4.5] . Average Velocity \(=\square\) (B) Find the instantaneous velocity when \(t=4\) \(=\square\) Instantaneous velocity \(=\square\)

Step-by-Step Solution

Answer

Average velocities: 2 m/s, 2.5 m/s, 4 m/s, 3.5 m/s. Instantaneous velocity at t = 4 is 3 m/s.

Step 1: Understand the average velocity

The average velocity over a time interval \text{[a, b]} is calculated using the formula: \[ \text{Average Velocity} = \frac{s(b) - s(a)}{b - a} \] where \( s(t) \) is the displacement function.

Step 2: Calculate displacement values

Evaluate the displacement function at the required points: \(s(t) = t^2 - 5t + 12 \) Find \( s \text{(3)}\), \( s \text{(4)} \), \( s \text{(3.5)} \), \( s \text{(5)} \), and \( s \text{(4.5)} \). \[ s(3) = 3^2 - 5(3) + 12 = 6 \] \[ s(4) = 4^2 - 5(4) + 12 = 8 \] \[ s(3.5) = (3.5)^2 - 5(3.5) + 12 = 6.75 \] \[ s(5) = 5^2 - 5(5) + 12 = 12 \] \[ s(4.5) = (4.5)^2 - 5(4.5) + 12 = 9.75 \]

Step 3: Calculate average velocity for [3, 4]

Using the displacement values from step 2, calculate the average velocity over the interval [3, 4]: \( \text{Average Velocity} = \frac{s(4) - s(3)}{4 - 3} \) \[ = \frac{8-6}{1} = 2 \text{ m/s} \]

Step 4: Calculate average velocity for [3.5, 4]

Using the displacement values from step 2, calculate the average velocity over the interval [3.5, 4]: \( \text{Average Velocity} = \frac{s(4) - s(3.5)}{4 - 3.5} \) \[= \frac{8 - 6.75}{0.5} = 2.5 \text{ m/s} \]

Step 5: Calculate average velocity for [4, 5]

Using the displacement values from step 2, calculate the average velocity over the interval [4, 5]: \( \text{Average Velocity} = \frac{s(5) - s(4)}{5 - 4} \) \[ = \frac{12 - 8}{1} = 4 \text{ m/s} \]

Step 6: Calculate average velocity for [4, 4.5]

Using the displacement values from step 2, calculate the average velocity over the interval [4, 4.5]: \( \text{Average Velocity} = \frac{s(4.5) - s(4)}{4.5 - 4} \) \[= \frac{9.75 - 8}{0.5} = 3.5 \text{ m/s} \]

Step 7: Find the derivative for instantaneous velocity

The instantaneous velocity is found by taking the derivative of the displacement function. Given \( s(t) = t^2 - 5t + 12 \), the derivative is: \[ s'(t) = 2t - 5 \]

Step 8: Calculate instantaneous velocity at t = 4

Substitute \( t = 4 \) into the derivative to find the instantaneous velocity. \[ s'(4) = 2(4) - 5 = 3 \text{ m/s} \]

Key Concepts

displacement function in kinematics

The displacement function in kinematics describes how the position of a particle changes over time. Displacement is a vector quantity, meaning it has both magnitude and direction. In our exercise, the displacement function is given by \(s(t) = t^2 - 5t + 12\). This equation tells us the position \(s\) of a particle at any time \(t\). To fully understand how the particle moves, we must evaluate this function at different times. When we plug in different values for \(t\), we can find the particle's position. For example, for \(t = 3\), the displacement \(s(3)\) is calculated by substituting \(t\) with 3 in the function: \[ s(3) = 3^2 - 5(3) + 12 = 6 \text{ meters} \] By doing this for more values, we map out the particle's journey:

• \( s(4) = 8 \text{ meters} \)
• \( s(3.5) = 6.75 \text{ meters} \)
• \( s(5) = 12 \text{ meters} \)
• \( s(4.5) = 9.75 \text{ meters} \)

average velocity calculation

Average velocity tells us how fast a particle's position changes over a time interval. It can be found using the formula: \[ \text{Average Velocity} = \frac{s(b) - s(a)}{b - a} \] Here, \( s(b)\) and \( s(a)\) represent the positions at times \( b \) and \( a \) respectively. The process involves:

• Finding the positions at the beginning and end of the interval using the displacement function.
• Subtracting the starting position from the ending position.
• Dividing the result by the length of the time interval.

{ For example, to find the average velocity between \(t = 3\) and \(t = 4\): \[\text{Average Velocity} = \frac{8 - 6}{4 - 3} = 2 \text{ m/s}\] }This tells us that between 3 and 4 seconds, the particle's position changes by 2 meters per second. Similarly, we can compute for other intervals in the given exercise:

• \([3.5, 4]\): \(\text{Average Velocity} = 2.5 \text{ m/s}\)
• \([4, 5]\): \(\text{Average Velocity} = 4 \text{ m/s}\)
• \([4, 4.5]\): \(\text{Average Velocity} = 3.5 \text{ m/s}\)

instantaneous velocity

Instantaneous velocity is the rate of change of displacement at a specific moment in time. It represents how fast a particle is moving at an exact time. To find it, we take the derivative of the displacement function. Mathematically, the instantaneous velocity \(v(t)\) is: \[ v(t) = s'(t)\] Where \(s'(t)\) is the derivative of the displacement function \(s(t)\). In our case: \[s(t) = t^2 - 5t + 12\] The derivative then is: \[s'(t) = 2t - 5\] To find the instantaneous velocity at \(t = 4\): we substitute 4 for \(t\) in the derivative: \(s'(4) = 2(4) - 5 = 3 \text{ m/s}\). This means that at 4 seconds, the particle is moving at 3 meters per second. Instantaneous velocity is crucial for understanding the exact speed of a particle at any given point in time, providing finer details about motion compared to average velocity.