Question

A particle leaves the origin at \(t=0\) with an initial velocity \(\overrightarrow{\mathbf{v}}_{0}=(3.6 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{i}}\). It experiences a constant acceleration \(\overrightarrow{\mathbf{a}}=-\left(1.2 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathbf{i}}-\left(1.4 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathbf{j}} .(a)\) At what time does the particle reach its maximum \(x\) coordinate? \((b)\) What is the velocity of the particle at this time? \((c)\) Where is the particle at this time?

Short answer

The particle reaches its maximum x-coordinate at \(t = 3\,s\). The velocity of the particle at this point is \(0\,\hat{i} - 4.2\,\hat{j}\, m/s\), and the position of the particle at this time is \(0\,\hat{i} - 6.3\,\hat{j}\,m\)

Step-by-step solution

Determine time at maximum x-coordinate (Solution to part (a))

The x-component of motion can be solved independently from the y-component as the two are perpendicular to each other. The equation for displacement in the x-direction under constant acceleration is \(x = v_0t + 0.5at^2\). Here, \(v_0\) is the x-component of the initial velocity, \(a\) is the x-component of the acceleration, and \(t\) is the time. To find the time at which the particle reaches its maximum x coordinate, we can set the derivative of the displacement function with respect to time equal to zero. Since \(x' = v_0 + at\) , when \(x'\) equals zero, we get \(t = -v_0/a\). Substituting \(v_0 = 3.6\,m/s\) and \(a = -1.2\,m/s^2\), we get \(t = 3\,s\).

Determine velocity at maximum x-coordinate (Solution to part (b))

The velocity in the x-direction at any time, \(t\), under constant acceleration can be found using the equation \(v = v_0 + at\). Substituting for \(v_0 = 3.6\,m/s\), \(a = -1.2\,m/s^2\) and \(t = 3\,s\) yields \(v_x = 0\,m/s\). The velocity in the y-direction can be found using the same equation, with \(v_0 = 0\), \(a = -1.4\, m/s^2\) and \(t = 3\,s\), providing \(v_y = -4.2\,m/s\). Thus, the velocity of the particle when it reaches max x-coordinate is \((0, -4.2)\,m/s\) or \(0\,\hat{i} - 4.2\,\hat{j}\,m/s\).

Determine the position at maximum x-coordinate (Solution to part (c))

We need to find the y-coordinate at \(t = 3\,s\). For the y component of the motion, \(v_0 = 0\) and \(a = -1.4\,m/s^2\). Using the equation \(y = v_0t + 0.5at^2\), and substituting for \(t = 3\,s\), \(v_0 = 0\), and \(a = -1.4\,m/s^2\) gives \(y = -6.3\,m\). Given that the maximum x-coordinate is obtained when x = 0, the particle's position at \(t = 3\,s\) is \((0, -6.3)\,m\) or \(0\,\hat{i} - 6.3\,\hat{j}\,m\).

Key concepts

Kinematics

Kinematics is the study of motion without considering forces. It helps us describe how objects move by looking at parameters like velocity, acceleration, and displacement. In the given problem, the particle's initial velocity and constant acceleration need to be analyzed to determine its motion. By understanding kinematic equations, we can predict past or future positions, velocities, or accelerations of the object in motion. We often use the equation \( x = v_0t + 0.5at^2 \) to predict the displacement over time when acceleration is constant. Kinematics forms the foundation for understanding more complex motions by allowing us to break down motion into simple components in straightforward coordinate directions.

Constant Acceleration

Constant acceleration occurs when the acceleration of an object does not change over time. In our problem, the particle has a constant acceleration in both the x and y directions:
- In the x-direction: \(-1.2 \, \text{m/s}^2\)
- In the y-direction: \(-1.4 \, \text{m/s}^2\)
Because acceleration is constant, we use specific kinematic equations to derive the motion parameters. For example, the velocity under constant acceleration can be found using \( v = v_0 + at \), where \( v_0 \) is the initial velocity and \( a \) is the constant acceleration. This principle allows us to calculate the velocity at any point, like when the particle reaches its maximum x-coordinate.

Velocity and Displacement Analysis

Analyzing velocity and displacement helps us understand how an object moves over time. In our problem, we first determine when the particle reaches its maximum x-coordinate by setting the derivative of the x displacement to zero:
\[ t = -\frac{v_0}{a} \].
This gives us the time when velocity in the x-direction becomes zero. For a particle with initial velocity \(3.6\, \text{m/s}\) and acceleration \(-1.2\, \text{m/s}^2\), \(t = 3\, \text{s}\).

To find its velocity at this time, we apply:
\( v = v_0 + at \).
In the x-direction, this results in zero velocity, while in the y-direction, it's \(-4.2\, \text{m/s}\).

Finally, determining displacement after \(3\, \text{s}\) gives us the positions of both x and y components to find the overall position of the particle.

Vectors in Physics

Vectors represent quantities that have both magnitude and direction, essential in physics for describing motion. Understanding vectors helps break down complex motions into understandable components. In our exercise, motion in two perpendicular directions—x and y—illustrates how vectors work.

• The initial velocity vector is \(\overrightarrow{v_0} = 3.6\, \text{m/s}\, \hat{i}\).
• Acceleration vector: \(\overrightarrow{a} = -1.2\, \text{m/s}^2 \, \hat{i} - 1.4\, \text{m/s}^2\, \hat{j}\).

We process vectors component-wise, analyzing x and y motions separately. This separation simplifies calculations, allowing us to apply kinematic equations to each direction independently. Mastery of vectors in physics equips students to visualize and solve more complex multidimensional problems.