Problem 19 - Fundamentals of Physics

Question

The acceleration of a particle moving only on a horizontal \(x y\) plane is given by \(\vec{a}=3 t \hat{i}+4 t \hat{j}\), where \(\vec{a}\) is in meters per secondsquared and \(t\) is in seconds. At \(t=0\), the position vector \(\vec{r}=(20.0 \mathrm{~m}) \hat{\mathrm{i}}+(40.0 \mathrm{~m}) \hat{\mathrm{j}}\) locates the particle, which then has the velocity vector \(\vec{v}=(5.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(2.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). At \(t=4.00 \mathrm{~s}\), what are (a) its position vector in unit-vector notation and (b) the angle between its direction of travel and the positive direction of the \(x\) axis?

Step-by-Step Solution

Answer

(a) \( \vec{r}(4) \approx 72 \hat{i} + 90.67 \hat{j} \). (b) \( \approx 49.4^\circ \) with the x-axis.

Step 1: Find the Velocity Function

The acceleration is given as \( \vec{a} = 3t \hat{i} + 4t \hat{j} \). To find the velocity, integrate the acceleration with respect to time:\[ \vec{v}(t) = \int \vec{a} \, dt = \int (3t \hat{i} + 4t \hat{j}) \, dt = \left( \frac{3t^2}{2} + C_1 \right) \hat{i} + \left( \frac{4t^2}{2} + C_2 \right) \hat{j} \].We use initial conditions that at \(t=0\), \(\vec{v} = 5\hat{i} + 2\hat{j}\). Substitute to find \(C_1\) and \(C_2\):1. \( \frac{3(0)^2}{2} + C_1 = 5 \Rightarrow C_1 = 5 \).2. \( \frac{4(0)^2}{2} + C_2 = 2 \Rightarrow C_2 = 2 \).Thus the velocity function is:\[ \vec{v}(t) = \left( \frac{3t^2}{2} + 5 \right) \hat{i} + \left( 2t^2 + 2 \right) \hat{j} \].

Step 2: Find the Position Function

Integrate the velocity function to find the position function:\[ \vec{r}(t) = \int \vec{v}(t) \, dt = \int \left( \frac{3t^2}{2} + 5 \right) \hat{i} + \left( 2t^2 + 2 \right) \hat{j} \, dt = \left( \frac{t^3}{2} + 5t + C_3 \right) \hat{i} + \left( \frac{2t^3}{3} + 2t + C_4 \right) \hat{j} \].Using initial conditions \( \vec{r}(0) = 20 \hat{i} + 40 \hat{j} \), solve for \(C_3\) and \(C_4\):1. \( \frac{(0)^3}{2} + 5*0 + C_3 = 20 \Rightarrow C_3 = 20 \).2. \( \frac{2(0)^3}{3} + 2*0 + C_4 = 40 \Rightarrow C_4 = 40 \).Thus the position function is:\[ \vec{r}(t) = \left( \frac{t^3}{2} + 5t + 20 \right) \hat{i} + \left( \frac{2t^3}{3} + 2t + 40 \right) \hat{j} \].

Step 3: Find the Position at t = 4 s

Substitute \( t = 4 \) into the position function:\[ \vec{r}(4) = \left( \frac{(4)^3}{2} + 5*4 + 20 \right) \hat{i} + \left( \frac{2(4)^3}{3} + 2*4 + 40 \right) \hat{j} \].Calculate each component:- \( \hat{i} \) component: \( \frac{64}{2} + 20 + 20 = 72 \, \text{m} \).- \( \hat{j} \) component: \( \frac{128}{3} + 8 + 40 \approx 90.67 \, \text{m} \).Thus, \( \vec{r}(4) \approx 72 \hat{i} + 90.67 \hat{j} \).

Step 4: Calculate the Direction Angle

At \( t = 4 \), find the velocity vector:\[ \vec{v}(4) = \left( \frac{3(4)^2}{2} + 5 \right) \hat{i} + \left( 2(4)^2 + 2 \right) \hat{j} = \left( 24 + 5 \right) \hat{i} + \left( 32 + 2 \right) \hat{j} = 29 \hat{i} + 34 \hat{j} \].The angle \( \theta \) with the positive \( x \)-axis is found using:\[ \theta = \tan^{-1} \left( \frac{34}{29} \right) \approx 49.4^\circ \].

Key Concepts

Velocity

Velocity is a critical concept in kinematics, representing the rate at which an object changes its position. It is a vector quantity, having both magnitude and direction. In the given problem, the velocity is initially given as \(\vec{v} = 5\hat{i} + 2\hat{j}\) m/s at \(t=0\). This shows the object is moving along both the x and y axes.
To find the velocity at any time \(t\), we integrate the acceleration function, \(\vec{a}(t)\), with respect to time. The integration process introduces integration constants, which can be solved using initial conditions. In this scenario, these conditions at \(t=0\) give \(C_1 = 5\) and \(C_2 = 2\).
So, the velocity function becomes \(\vec{v}(t) = \left(\frac{3t^2}{2} + 5\right) \hat{i} + \left(2t^2 + 2\right) \hat{j}\). This equation allows you to calculate the velocity at any time \(t\), understanding how the object's speed and direction evolve.

Acceleration

Acceleration is the rate at which velocity changes with time. It provides information about how an object speeds up, slows down, or changes direction. In unit vector notation, it's specified as \( \vec{a} = 3t \hat{i} + 4t \hat{j} \), which means the particle's acceleration vector changes linearly with time.
Since this is a vector quantity, each component contributes differently to the particle's acceleration:

• In the \(\hat{i}\) direction, the component is \(3t\), implying that with every second, the acceleration in the x-direction increases by 3 meters per second squared.
• In the \(\hat{j}\) direction, the component is \(4t\), indicating a faster increase in acceleration in the y-direction, compared to the x-direction.

Understanding these contributions is essential to predict the particle's future velocity and position accurately.

Position Vector

The position vector describes the location of a particle in a given space, considering both the magnitude and direction. Here, the initial position vector at \(t=0\) is \(\vec{r} = 20 \hat{i} + 40 \hat{j}\) meters, indicating the particle starts 20 meters along the x-axis and 40 meters up the y-axis.
To determine how the position changes over time, we integrate the velocity function, yielding the position function \(\vec{r}(t) = \left(\frac{t^3}{2} + 5t + 20\right) \hat{i} + \left(\frac{2t^3}{3} + 2t + 40\right) \hat{j}\). Using this function, you can find where the particle is at any given time, like at \(t=4\) seconds, showing it moves to approximately \(72 \hat{i} + 90.67 \hat{j}\).
This way, the position vector's analysis assists in visualizing the trajectory of the particle across the \(xy\) plane.

Unit Vector Notation

Unit vector notation is a helpful tool in physics and engineering to represent vectors accurately. A unit vector indicates direction and is generally denoted with symbols like \(\hat{i}, \hat{j},\) and \(\hat{k}\) for 3D space, which point along the x, y, and z axes, respectively.
In our exercise:

• \(\hat{i}\) represents motion or dimension in the x-axis direction.
• \(\hat{j}\) symbolizes the y-axis direction.
• When necessary, \(\hat{k}\) would correspond to the z-axis, although it's not present in this 2D problem.

Using unit vectors allows each component to be dealt with separately, simplifying the calculation processes necessary for determining resultant vectors like velocity \(\vec{v}(t)\) and position \(\vec{r}(t)\). This method makes complex vector operations more intuitive and provides clarity in handling multidirectional movements.