Problem 25 - Physics for Scientists and Engineers with Modern Physics

Question

(II) Suppose the position of an object is given by \(\vec{\mathbf{r}}=\left(3.0 t^{2} \hat{\mathbf{i}}-6.0 t^{3} \hat{\mathbf{j}}\right) \mathrm{m}\) (a) Determine its velocity \(\vec{\mathbf{v}}\) and acceleration \(\vec{a},\) as a function of time. (b) Determine \(\vec{\mathbf{r}}\) and \(\vec{\mathbf{v}}\) at time \(t=2.5 \mathrm{s} .\)

Step-by-Step Solution

Answer

Velocity: \( \vec{\mathbf{v}} = 6.0t \hat{\mathbf{i}} - 18.0t^2 \hat{\mathbf{j}} \); Acceleration: \( \vec{\mathbf{a}} = 6.0 \hat{\mathbf{i}} - 36.0t \hat{\mathbf{j}} \). At \( t=2.5 \): \( \vec{\mathbf{r}} = 18.75 \hat{\mathbf{i}} - 93.75 \hat{\mathbf{j}} \), \( \vec{\mathbf{v}} = 15.0 \hat{\mathbf{i}} - 112.5 \hat{\mathbf{j}} \).

Step 1: Understand the position vector

The position vector of the object is given by \( \vec{\mathbf{r}} = (3.0t^2 \hat{\mathbf{i}} - 6.0t^3 \hat{\mathbf{j}}) \mathrm{m} \). This means that the x-component of the position is \( 3.0t^2 \) and the y-component is \( -6.0t^3 \). We need to differentiate these components to find velocity and acceleration.

Step 2: Differentiate to find velocity

The velocity \( \vec{\mathbf{v}} \) is the first derivative of the position vector \( \vec{\mathbf{r}} \) with respect to time \( t \). So, calculate:\[ \vec{\mathbf{v}} = \frac{d}{dt}(3.0t^2 \hat{\mathbf{i}} - 6.0t^3 \hat{\mathbf{j}}) = (\frac{d}{dt}(3.0t^2) \hat{\mathbf{i}}) + (\frac{d}{dt}(-6.0t^3) \hat{\mathbf{j}}) \]\[ \vec{\mathbf{v}} = (6.0t \hat{\mathbf{i}} - 18.0t^2 \hat{\mathbf{j}}) \mathrm{m/s} \]

Step 3: Differentiate to find acceleration

The acceleration \( \vec{\mathbf{a}} \) is the first derivative of the velocity vector \( \vec{\mathbf{v}} \) with respect to time \( t \). Differentiate \( \vec{\mathbf{v}} \):\[ \vec{\mathbf{a}} = \frac{d}{dt}(6.0t \hat{\mathbf{i}} - 18.0t^2 \hat{\mathbf{j}}) \] \[ \vec{\mathbf{a}} = (6.0 \hat{\mathbf{i}} - 36.0t \hat{\mathbf{j}}) \mathrm{m/s^2} \]

Step 4: Evaluate position at \( t = 2.5 \) s

To find the position \( \vec{\mathbf{r}} \) at \( t = 2.5 \) s, substitute \( t = 2.5 \) into the position equation:\[ \vec{\mathbf{r}} = 3.0(2.5)^2 \hat{\mathbf{i}} - 6.0(2.5)^3 \hat{\mathbf{j}} \]\[ \vec{\mathbf{r}} = 18.75 \hat{\mathbf{i}} - 93.75 \hat{\mathbf{j}} \mathrm{m} \]

Step 5: Evaluate velocity at \( t = 2.5 \) s

To find velocity \( \vec{\mathbf{v}} \) at \( t = 2.5 \) s, substitute \( t = 2.5 \) into the velocity equation:\[ \vec{\mathbf{v}} = 6.0(2.5) \hat{\mathbf{i}} - 18.0(2.5)^2 \hat{\mathbf{j}} \]\[ \vec{\mathbf{v}} = 15.0 \hat{\mathbf{i}} - 112.5 \hat{\mathbf{j}} \mathrm{m/s} \]

Key Concepts

Position Vector

The position vector, often denoted as \( \vec{\mathbf{r}} \), is a fundamental concept in physics, representing the location of an object in space relative to an origin. In this exercise, the position vector is given as \( \vec{\mathbf{r}} = (3.0t^2 \hat{\mathbf{i}} - 6.0t^3 \hat{\mathbf{j}}) \mathrm{m} \).
This expression tells us how the object's position changes over time along the x and y axes:

• X-component: \( 3.0t^2 \hat{\mathbf{i}} \) indicates that the position in the x-direction depends on \( t^2 \).
• Y-component: \( -6.0t^3 \hat{\mathbf{j}} \) means the position in the y-direction depends on \( t^3 \).

This vector form illustrates that motion in the x-direction and y-direction can be described independently. By analyzing the position vector, one can determine how an object moves in a two-dimensional plane over time.

Velocity

Velocity \( \vec{\mathbf{v}} \) is the rate of change of position with respect to time, essentially describing how quickly an object changes its position. In simpler terms, velocity is how fast something is moving and in which direction. To find the velocity, we differentiate the position vector with respect to time \( t \).
This gives us: \[ \vec{\mathbf{v}} = \frac{d}{dt}(3.0t^2 \hat{\mathbf{i}} - 6.0t^3 \hat{\mathbf{j}}) = (6.0t \hat{\mathbf{i}} - 18.0t^2 \hat{\mathbf{j}}) \mathrm{m/s} \]Each component of the velocity vector shows:

• X-component: \( 6.0t \hat{\mathbf{i}} \), meaning as time increases, velocity in the x-direction increases linearly with time.
• Y-component: \( -18.0t^2 \hat{\mathbf{j}} \), indicating a quadratic change in velocity for the y-direction.

Thus, the velocity vector provides both speed and direction of an object's motion.

Acceleration

Acceleration \( \vec{\mathbf{a}} \) is the rate of change of velocity with respect to time, showing how quickly an object is speeding up or slowing down. To calculate acceleration, we differentiate the velocity vector with respect to time \( t \).
This results in:\[ \vec{\mathbf{a}} = \frac{d}{dt}(6.0t \hat{\mathbf{i}} - 18.0t^2 \hat{\mathbf{j}}) = (6.0 \hat{\mathbf{i}} - 36.0t \hat{\mathbf{j}}) \mathrm{m/s^2} \]
The components of acceleration are:

• X-component: A constant acceleration of \( 6.0 \hat{\mathbf{i}} \), suggesting constant acceleration in the x-direction.
• Y-component: \( -36.0t \hat{\mathbf{j}} \), showing linear time-dependence that implies acceleration in the y-direction changes with time.

Understanding acceleration is crucial in predicting future motion, as it tells us not just the speed of an object but how that speed changes over time.

Differentiation

Differentiation is a mathematical process used to find the rate at which a quantity changes, known as a derivative. In the context of mechanics, differentiation helps in analyzing motion by finding velocity from the position vector and acceleration from the velocity vector.
To perform differentiation, apply the derivative with respect to time \( t \):

• By differentiating the position vector \( \vec{\mathbf{r}} \), we obtain the velocity vector \( \vec{\mathbf{v}} \).
• By further differentiating the velocity vector \( \vec{\mathbf{v}} \), we derive the acceleration vector \( \vec{\mathbf{a}} \).

These differentiated quantities allow us to understand and describe the dynamics of an object precisely, providing a clear picture of how its motion evolves over time. Differentiation thus serves as a powerful tool in classical mechanics for analyzing changes in motion.