Problem 13 - College Physics

Question

A particle moves at a speed of \(3.0 \mathrm{~m} / \mathrm{s}\) in the \(+x\) -direction. Upon reaching the origin, the particle receives a continuous constant acceleration of \(0.75 \mathrm{~m} / \mathrm{s}^{2}\) in the \(-y\) -direction. What is the position of the particle 4.0 s later?

Step-by-Step Solution

Answer

The position of the particle is \((12.0 \ \mathrm{m}, -6.0 \ \mathrm{m})\) 4.0 s later.

Step 1: Initial Conditions

First, note the initial conditions for the particle's motion. The initial velocity of the particle is \(v_{x0} = 3.0 \ \mathrm{m/s}\) in the \(+x\) direction, and the initial velocity in the \(y\) direction is \(v_{y0} = 0 \ \mathrm{m/s}\) since the particle receives acceleration at the origin.

Step 2: Determine Motion in the x-direction

Since there is no acceleration in the \(x\) direction, the velocity remains constant at \(v_x = 3.0 \ \mathrm{m/s}\). Use the equation \(x = v_{x0} \cdot t\) to find the position in the \(x\) direction:\[ x = 3.0 \ \mathrm{m/s} \times 4.0 \ \mathrm{s} = 12.0 \ \mathrm{m} \]

Step 3: Determine Motion in the y-direction

In the \(y\) direction, the particle has an initial velocity \(v_{y0} = 0 \ \mathrm{m/s}\), but there is a constant acceleration \(a_y = -0.75 \ \mathrm{m/s}^2\). Use the equation \(y = v_{y0} \cdot t + \frac{1}{2} a_y t^2\):\[ y = 0 \cdot 4.0 + \frac{1}{2} \times (-0.75) \times 4.0^2 = -6.0 \ \mathrm{m} \]

Step 4: Combine x and y positions

Combine the positions to find the final position vector of the particle. The position at \(t = 4.0 \ \mathrm{s}\) is:\[ \vec{r}(t) = \langle x, y \rangle = \langle 12.0, -6.0 \rangle \] Therefore, at \(4.0 \ \mathrm{s}\), the position of the particle is \(12.0 \ \mathrm{m}\) in the \(x\) direction and \(-6.0 \ \mathrm{m}\) in the \(y\) direction.

Key Concepts

Initial Velocity

Initial velocity is the speed and direction of an object at the beginning of a time interval. In kinematics, understanding the initial velocity allows us to predict how an object will move over time.
For the exercise at hand, the particle starts with an initial velocity of \(v_{x0} = 3.0 \, \mathrm{m/s}\) along the positive \(x\)-axis. This means that initially, it moves steadily to the right.
Meanwhile, the particle's initial velocity in the \(y\)-direction is \(v_{y0} = 0 \, \mathrm{m/s}\), indicating no initial motion up or down. This is key because, after this point, changes only occur in the \(y\)-direction due to acceleration.

Constant Acceleration

Constant acceleration occurs when an object's velocity changes at a steady rate over time. This is a crucial concept because it allows us to apply simple equations to calculate future positions or speeds.
In this exercise, the particle experiences a constant acceleration of \(-0.75 \, \mathrm{m/s}^2\) in the negative \(y\)-direction. This steady increase in speed downwards means that although our particle had no initial \(y\)-motion, it will progressively move downwards over time. Constant acceleration is essential for calculating its position after a set duration.

Position Vector

A position vector gives a complete description of an object's location in space by combining both direction and magnitude. It essentially pinpoints where an object is at a specific moment in time.
The position vector in our problem helps us understand where the particle is after 4 seconds. It combines the movements in both \(x\) and \(y\) directions to \(\langle 12.0, -6.0 \rangle\). This vector tells us that the particle is 12 meters to the right and 6 meters downward from the origin.
Position vectors are a powerful tool in visualizing an object's current position in a two or three-dimensional plane.

Motion Equations

Motion equations in kinematics help us calculate where an object will be at a future time given its initial conditions and accelerations. They allow us to predict the paths of moving objects.
For movement in the \(x\)-direction, since there's no acceleration, we use the simple equation \(x = v_{x0} \cdot t\), which yields \(12.0 \, \mathrm{m}\).
For the \(y\)-direction, we use \(y = v_{y0} \cdot t + \frac{1}{2} a_y t^2\) to find that the particle moves to \(-6.0 \, \mathrm{m}\) after 4 seconds.
These equations enable the breakdown of motion into easily manageable parts, and they are essential for solving kinematics problems.