Problem 17 - Physics for Scientists and Engineers with Modern Physics

Question

The position of a particular particle as a function of time is given by \(\overrightarrow{\mathbf{r}}=\left(9.60 t \hat{\mathbf{i}}+8.85 \hat{\mathbf{j}}-1.00 t^{2} \hat{\mathbf{k}}\right) \mathrm{m} .\) Determine the particle's velocity and acceleration as a function of time.

Step-by-Step Solution

Answer

Velocity: \( (9.60 \hat{i} - 2.00t \hat{k}) \) m/s, Acceleration: \( -2.00 \hat{k} \) m/s².

Step 1: Understand the Position Function

The position of the particle, \( \vec{r} \), is given by \( \vec{r} = (9.60t \hat{i} + 8.85 \hat{j} - 1.00t^2 \hat{k}) \, \text{m} \). This tells us that the position depends on time \( t \) in a vector form with components along \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \).

Step 2: Determine the Velocity Function

Velocity is the first derivative of the position with respect to time. Therefore, we find \( \vec{v} = \frac{d\vec{r}}{dt} \). Derivative of \( 9.60t \) with respect to \( t \) is \( 9.60 \), derivative of \( 8.85 \) is \( 0 \), and \( \frac{d}{dt}(-1.00t^2) = -2.00t \). So, \( \vec{v}(t) = (9.60 \hat{i} + 0 \hat{j} - 2.00t \hat{k}) \) m/s.

Step 3: Determine the Acceleration Function

Acceleration is the first derivative of the velocity with respect to time. So, find \( \vec{a} = \frac{d\vec{v}}{dt} \). Derivative of \( 9.60 \) is \( 0 \), derivative of \( 0 \) is \( 0 \), and \( \frac{d}{dt}(-2.00t) = -2.00 \). Thus, \( \vec{a}(t) = (0 \hat{i} + 0 \hat{j} - 2.00 \hat{k}) \) m/s².

Key Concepts

Velocity

Velocity is a crucial concept in kinematics. It is the rate of change of an object's position as a function of time. To find the velocity from a position function, like the one given in our exercise, you need to take the derivative of the position function with respect to time.

This means for a position function such as \( \vec{r} = (9.60t \hat{i} + 8.85 \hat{j} - 1.00t^2 \hat{k}) \text{ m} \), the velocity function \( \vec{v}(t) \) is obtained by differentiating each component with respect to time.

• The derivative of \( 9.60t \) along \( \hat{i} \) is \( 9.60 \).
• Along \( \hat{j} \), since \( 8.85 \) is a constant, its derivative is \( 0 \).
• The \( \hat{k} \) component \( -1.00t^2 \) becomes \( -2.00t \) after differentiation.

This gives the velocity vector: \( \vec{v}(t) = (9.60 \hat{i} + 0 \hat{j} - 2.00t \hat{k}) \) m/s. Velocity is itself a vector quantity, possessing both magnitude and direction, which makes it different from speed, a scalar quantity.

Acceleration

Acceleration reflects how quickly an object's velocity changes over time. It is another fundamental component in understanding motion and is especially important when forces are involved. The acceleration function is derived from the velocity function by differentiating it with respect to time.

In our solution, starting from the velocity \( \vec{v}(t) = (9.60 \hat{i} + 0 \hat{j} - 2.00t \hat{k}) \) m/s, we differentiate:

• The constant \( 9.60 \) along \( \hat{i} \) results in \( 0 \) since constants do not change with time.
• The component \( 0 \) along \( \hat{j} \) remains \( 0 \).
• Finally, \( -2.00t \) along \( \hat{k} \) becomes \( -2.00 \).

The acceleration vector thus is \( \vec{a}(t) = (0 \hat{i} + 0 \hat{j} - 2.00 \hat{k}) \) m/s². It's crucial to understand that just like velocity, acceleration also has a direction and a magnitude, providing a complete description of the change in velocity.

Position Function

The position function describes the location of a particle in space at any given time. In this exercise, the position function \( \vec{r} = (9.60t \hat{i} + 8.85 \hat{j} - 1.00t^2 \hat{k}) \text{ m} \) gives us a vector in 3D space showing the position's dependence on time.

This function is an example of a vector function with components:

• \( 9.60t \hat{i} \) indicates the position along the x-axis depends linearly on time with a coefficient of \( 9.60 \).
• \( 8.85 \hat{j} \) indicates a constant position along the y-axis.
• \( -1.00t^2 \hat{k} \) indicates the position along the z-axis follows a quadratic relation with time, showing that it accelerates over time.

Understanding how each component of the vector behaves is crucial for predicting the future position of the particle. The position function allows us to visualize the motion path by plotting it over a time interval, providing insights into the particle's movement.