Question

An electron, starting from rest and moving with a constant acceleration, travels \(1.0 \mathrm{~cm}\) in \(2.0 \mathrm{~ms}\). What is the magnitude of this acceleration? a) \(25 \mathrm{~km} / \mathrm{s}^{2}\) b) \(20 \mathrm{~km} / \mathrm{s}^{2}\) c) \(15 \mathrm{~km} / \mathrm{s}^{2}\) d) \(10 \mathrm{~km} / \mathrm{s}^{2}\) e) \(5.0 \mathrm{~km} / \mathrm{s}^{2}\)

Short answer

a) 0.05 km/s² b) 0.5 km/s² c) 1.0 km/s² d) 2.5 km/s² e) 5.0 km/s² Answer: e) 5.0 km/s²

Step-by-step solution

Identify known and unknown variables

The known variables: Initial velocity (u) = \(0~\text{m/s}\) Distance (s) = \(1.0~\text{cm}\) = \(1.0\times10^{-2}~\text{m}\) Time (t) = \(2.0~\text{ms}\) = \(2.0\times10^{-3}~\text{s}\) The unknown variable: Acceleration (a) = ?

Use the equation of motion to solve for acceleration

The equation of motion is: \(s = ut + \frac{1}{2}at^2\) Substitute the known values into the equation: \(1.0\times10^{-2}~\text{m} = (0~\text{m/s})(2.0\times10^{-3}~\text{s}) + \frac{1}{2}a(2.0\times10^{-3}~\text{s})^2\)

Solve for acceleration

The equation simplifies to: \(1.0\times10^{-2}~\text{m} = \frac{1}{2}a(4.0\times10^{-6}~\text{s}^2)\) Now, we can solve for acceleration (a): \(a = \frac{2(1.0\times10^{-2}~\text{m})}{4.0\times10^{-6}~\text{s}^2} = 5000 ~\mathrm{m}/\mathrm{s}^2\)

Convert the acceleration to km/s^2

To convert the acceleration unit from m/s² to km/s², divide the value by 1000: \(a = \frac{5000 ~\mathrm{m}/\mathrm{s}^2}{1000} = 5.0 ~\mathrm{km}/\mathrm{s}^2\)

Compare the result with the given options

So, the magnitude of acceleration is \(5.0 ~\mathrm{km}/\mathrm{s}^2\). Comparing this result with the given options, we find that option e) is the correct answer. Answer: e) \(5.0 \mathrm{~km} / \mathrm{s}^{2}\)

Key concepts

Constant Acceleration

Understanding constant acceleration is crucial in solving many physics problems because it simplifies motion analysis. Acceleration is defined as the rate of change of velocity per unit time. Constant acceleration means that the rate of change of velocity is uniform over time. This is a critical point, especially when an object, like the electron in our exercise, starts from rest. In such scenarios, if the acceleration is constant, we can apply the equations of motion derived specifically for uniformly accelerating objects. These equations provide a reliable method to calculate unknown variables—such as displacement, final velocity, or time—if we know the initial conditions and the acceleration value. In our example, the electron has an acceleration that uniformly increases its velocity from zero to a certain value over a given time and distance.

Motion Kinematics

Motion kinematics is the study of motion without considering the forces that cause such motion. In kinematics, we deal with variables like displacement, velocity, acceleration, and time. The equations of motion used in kinematics to describe the characteristics of a moving object under constant acceleration are essential tools.

For example, the second equation of motion, which is \(s = ut + \frac{1}{2}at^2\), was used in the given problem. Here, \(s\) represents the distance traveled, \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time. Since the problem provides the distance (\(s\)) and time (\(t\)), and we know the electron starts from rest (\(u = 0\ m/s\)), the equation allowed us to isolate and solve for the acceleration (\(a\)).

Physics Problem Solving

Problem-solving in physics is a systematic process that involves understanding the problem, visualizing it, breaking it down to basic principles, applying relevant formulas, and carrying out the required computations. The approach to the electron's acceleration problem highlights this method.

First, we identified the knowns and unknowns. Then we applied a relevant equation of motion, reflecting a conceptual understanding of the problem. By substituting the known values into the equation, we were able to algebraically solve for the unknown. This step-by-step process not only helps in finding the correct answer but also reinforces the understanding of the underlying physics concepts.

Unit Conversion

In physics, proper unit conversion is fundamental to getting correct results. The International System of Units (SI) provides a standard for measuring different physical quantities. However, many physics problems, like the one discussed, require conversions to alternative units for an appropriate comparison with given options.

After solving for acceleration in SI units (meters per second squared), we needed to convert to kilometers per second squared (\(\mathrm{km}/\mathrm{s}^2\)) to match the answer choices. This involved dividing the calculated acceleration by 1,000 since 1 kilometer equals 1,000 meters. This type of unit conversion is often encountered in physics problems and is an essential skill for students to master.