Question

In a three-dimensional motion, the \(x-, y-\), and \(z\) coordinates of the object as a function of time are given by \(x(t)=\frac{\sqrt{2}}{2} t, \quad y(t)=\frac{\sqrt{2}}{2} t,\) and \(z(t)=-4.9 t^{2}+\sqrt{3} t\) Describe the motion and the trajectory of the object in an \(x y z\) coordinate system.

Short answer

Answer: The motion of the object is linear along the diagonal x=y in the xy-plane, and it experiences constant downward acceleration in the z-direction. The trajectory equation relating x, y, and z is: \(z=-4.9\left(\frac{2x}{\sqrt{2}}\right)^2+\sqrt{3}\left(\frac{2x}{\sqrt{2}}\right)\) where \(x = y\).

Step-by-step solution

Determine the motion in the x direction

The equation for the x-coordinate is given as \(x(t)=\frac{\sqrt{2}}{2}t\). This shows that the x-coordinate increases linearly with time. As time goes on, the object moves further in the positive x-direction.

Determine the motion in the y direction

The equation for the y-coordinate is given as \(y(t)=\frac{\sqrt{2}}{2}t\). This means that the y-coordinate also increases linearly with time, at the same rate as the x-coordinate. As time goes on, the object moves further in the positive y-direction.

Determine the motion in the z direction

The equation for the z-coordinate is given as \(z(t)=-4.9t^2+\sqrt{3}t\). This equation has a quadratic term, so the object's motion in the z-direction won't be linear and will have some acceleration. The acceleration is given by the second derivative of this equation, which is -9.8 (a constant). The object experiences a constant downward acceleration (negative z-direction) while moving upward initially, due to the positive initial velocity.

Eliminate the time variable to find the trajectory

Since both \(x(t)\) and \(y(t)\) have the same dependency on time (\(\frac{\sqrt{2}}{2}t\)), we can say that \(x = y\). In other words, the object moves along the diagonal of the xy-plane. We can write the trajectory in the z-direction as a function of x (or y) by using our previous expression for x(t): \(t=\frac{2x}{\sqrt{2}}\). Substituting this expression for time into \(z(t)\), we obtain: \(z=\left(-4.9\left(\frac{2x}{\sqrt{2}}\right)^2+\sqrt{3}\left(\frac{2x}{\sqrt{2}}\right)\right)\).

Describe the motion and the trajectory

In summary, the object moves linearly along the diagonal x=y in the xy-plane and experiences constant downward acceleration in the z-direction. The trajectory equation relating x, y, and z is: \(z=-4.9\left(\frac{2x}{\sqrt{2}}\right)^2+\sqrt{3}\left(\frac{2x}{\sqrt{2}}\right)\) where \(x = y\) This equation describes the path the object follows in the xyz-coordinate system.

Key concepts

Equations of Motion

In 3D motion, equations of motion give us the mathematical relationship between position, velocity, and time. For this particular exercise, we have functions for each spatial dimension:

• X-direction: \(x(t) = \frac{\sqrt{2}}{2} t\)
• Y-direction: \(y(t) = \frac{\sqrt{2}}{2} t\)
• Z-direction: \(z(t) = -4.9 t^{2} + \sqrt{3} t\)

These equations help us understand how the object's position changes over time in each direction. The x and y equations show linear motion, indicating a constant velocity in those directions. The z equation, which includes a quadratic term, suggests a more complex motion involving acceleration. This setup is crucial for predicting and analyzing motion in real-world problems.

Trajectory Analysis

Trajectory refers to the path that a moving object follows through space. To determine the object's trajectory in this exercise, it's important to analyze both the linear and quadratic components combined from the motion equations. Here, we observe:

• The object moves linearly in both the x and y directions since \(x = y\).
• In the z direction, the object's motion is affected by gravity, as observed by the quadratic nature of \(z(t)\).

To find a single expression that describes the entire path of the object, we eliminate time \(t\) from these equations, resulting in a function of \(z\) with respect to \(x\) (or \(y\)). The resultant trajectory equation is:\[z = -4.9\left(\frac{2x}{\sqrt{2}}\right)^2 + \sqrt{3}\left(\frac{2x}{\sqrt{2}}\right)\]This equation captures the curved path of the object in the 3D coordinate system.

Coordinate System

The coordinate system in this context is an xyz-system, where movement is described across three orthogonal axes — x, y, and z. Understanding this system is essential to frame and interpret the object's motion:

• The x and y axes describe the object's horizontal position and are responsible for portraying the linear path along the plane.
• The z axis adds depth to this, detailing the object's vertical (or height) position over time.

Knowing how these axes interact is critical to visualize the object's 3D path clearly. The key takeaway is that linear increments along the x and y axes combine with the quadratic motion along the z axis to paint a full picture of how the object traverses through space.

Linear and Quadratic Motion

Linear motion occurs when an object travels in a straight line with constant velocity. Here, our equations \(x(t) = \frac{\sqrt{2}}{2} t\) and \(y(t) = \frac{\sqrt{2}}{2} t\) highlight the object's constant speed in these two dimensions. Such motion is predictable and easy to graph.
Conversely, quadratic motion involves acceleration, as seen in the equation for z: \(z(t) = -4.9 t^{2} + \sqrt{3} t\). The presence of a square term (\(-4.9 t^{2}\)) indicates that the z-motion is not only directed but also varying in speed — initially increasing due to gravity.
To summarize, examining both the linear and quadratic aspects of motion allows for a comprehensive understanding of how different forces affect a moving object in three-dimensional space. This analysis is helpful for everything from simple physics problems to complex engineering tasks.