Question

In each of the following problems, \(z\) represents the displacement of a particle from the origin. Find (as functions of \(t\) ) its speed and the magnitude of its acceleration, and describe the motion. \(z=(1+i) t-(2+i)(1-t) . \quad\) Hint: Show that the particle moves along a straight line through the points \((1+i)\) and \((-2-i)\).

Short answer

The speed of the particle is \(\sqrt{13}\) and the magnitude of the acceleration is 0. The motion is along a straight line through points \(1+i\) and \(-2-i\).

Step-by-step solution

- Simplify the expression for z

Start by simplifying the given expression for the displacement: \(z = (1 + i)t - (2 + i)(1 - t)\)

- Distribute and Collect Like Terms

Distribute the terms and combine like terms: \(z = (1 + i)t - (2 + i) + (2 + i)t\)

- Combine like terms

Combine the like terms by grouping all terms that involve \(t\): \(z = (1 + i)t + (2 + i)t - (2 + i)\) Next, factor out \(t\) from the terms: \(z = [(1+i) + (2+i)]t - (2 + i)\) which simplifies to \(z = (3 + 2i)t - (2 + i)\) So the simplified expression is \(z = (3 + 2i)t - 2 - i\)

- Identify the form

Rewrite the expression to explicitly see the form of a straight line. Given that the general form is \(z = at + b\) where \(a = 3 + 2i\) and \(b = -2 - i\), we can interpret that \(a\) represents the direction vector and \(b\) the starting point.

- Compute Speed

The speed is the magnitude of the direction vector: \(|a| = \big|3+2i\big| = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}\). Hence, the speed of the particle is \(\sqrt{13}\).

- Find Acceleration Magnitude

Since the acceleration term involves the second derivative of displacement with respect to time and \(z = (3 + 2i)t - (2 + i)\) is a linear function, the second derivative is zero. Therefore, the magnitude of acceleration is 0.

- Describe the Motion

Since the particle moves along the line defined by \(z = (3+2i)t -2 - i\), it will move in a straight line through the points \(1 + i\) and \(-2 - i\) with constant speed \(\sqrt{13}\) and no acceleration (as the acceleration is 0).

Key concepts

speed calculation

In this exercise, the particle's displacement is given by the complex function \(z = (1 + i)t - (2 + i)(1 - t)\). To compute the speed, we need to find the magnitude of the direction vector. After simplifying the equation, we found that the simplified form is \(z = (3 + 2i)t - 2 - i\). The direction vector's magnitude is computed by taking the square root of the sum of the squares of the real and imaginary components. In LaTeX, it’s written as:
\[|a| = |3+2i| = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}\]. Thus, the particle's speed is \(\sqrt{13}\). Speed is a measure of how fast an object moves and is always positive. In this case, it tells us the constant rate at which the particle moves along its path.

acceleration magnitude

The acceleration of a particle in motion describes the rate at which its velocity changes. It’s the derivative of the velocity or the second derivative of the displacement. For our equation \(z = (3 + 2i)t - 2 - i\), we see it’s a linear function concerning time \(t\). The first derivative of this function is the direction vector \(3 + 2i\), which represents the velocity:
\[v = \frac{dz}{dt} = 3 + 2i\].
Taking the derivative of the velocity, the second derivative, gives us the acceleration:
\[a = \frac{d^2z}{dt^2} = 0\]. This zero acceleration tells us that the particle’s velocity remains constant over time, thus moving with a constant speed along the path.

linear motion analysis

Linear motion occurs when an object travels along a straight path. In this exercise, after simplifying the displacement function, we obtained the expression \(z = (3 + 2i)t - 2 - i\). This form resembles the equation of a straight line in complex numbers, which confirms that the particle moves in a straight line. The vector \(3 + 2i\) is the direction vector, it points from the initial point towards the direction of motion. The initial point or starting position is given by the constant term \(-2 - i\). Hence the particle moves linearly from \((1+i)\) to \((-2-i)\) with time. Linear motion analysis in physics is crucial because it simplifies understanding the system’s dynamics under constant velocity and no acceleration.

complex numbers

Complex numbers consist of a real part and an imaginary part and are often written in the form \(a + bi\). Here, \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit, where \(i^2 = -1\). In our context, complex numbers represent the displacement of a particle in a 2D space where the real part denotes the x-coordinate and the imaginary part the y-coordinate. By simplifying the given equation to \(z = (3 + 2i)t - 2 - i\), we see how complex numbers are used to describe the motion of the particle along a straight line. This complex displacement combines real and imaginary parts to depict motion in a plane. Understanding complex numbers is fundamental in solving problems involving 2D motion and helps visualize the trajectory (or path) of moving objects.