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Chapter 3: Problem 24

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

Expert verified
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

01

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.
02

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.
03

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.
04

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: \(z=\left(-4.9\left(\frac{2x}{\sqrt{2}}\right)^2+\sqrt{3}\left(\frac{2x}{\sqrt{2}}\right)\right)\).
05

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.

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