Problem 37 - Physics for Scientists and Engineers with Modern Physics

Question

$$ \begin{array}{l}{\text { (II) Three students derive the following cquations in which }} \\ {x \text { refers to distance traveled, } v \text { the specd, } a \text { the acceleration }} \\ {\left(m / s^{2}\right), t \text { the time, and the subscript zero }(a) \text { means a quantity }}\end{array} $$ $$\begin{array}{l}{\text { at time } t=0 : \text { (a) } x=u t^{2}+2 a t,(b) x=v_{0} t+\frac{1}{2} a t^{2}, \text { and }} \\ {(c) \quad x=v_{0} t+2 a t^{2} \text { . Which of these could possibly be }} \\ {\text { correct according to a dimensional check? }}\end{array} $$

Step-by-Step Solution

Answer

Equations (b) and (c) are correct according to dimensional analysis.

Step 1: Understand Dimensional Analysis

Dimensional analysis involves checking that every term in an equation has the same unit or dimension. We will use the following dimensions: \(x\) has the dimension of length \([L]\), \(v_0\) has the dimension of speed \([LT^{-1}]\), \(a\) has the dimension of acceleration \([LT^{-2}]\), and \(t\) has the dimension of time \([T]\).

Step 2: Check Equation (a)

Equation (a) is \(x = u t^2 + 2a t\). The dimensional analysis is as follows:- Term \(u t^2\): \(u\) is unknown, dimension not given, so we can't check it correctly without knowing its dimension.- Term \(2a t\): Dimension of \(a\) is \([LT^{-2}]\) and \(t\) is \([T]\), resulting in \([LT^{-1}]\), which doesn't match \([L]\).Therefore, equation (a) fails dimensional analysis.

Step 3: Check Equation (b)

Equation (b) is \(x = v_0 t + \frac{1}{2} a t^2\). The dimensional analysis is as follows:- Term \(v_0 t\): Dimension of \(v_0\) is \([LT^{-1}]\) and \(t\) is \([T]\), resulting in \([L]\).- Term \(\frac{1}{2} a t^2\): Dimension of \(a\) is \([LT^{-2}]\) and \(t^2\) is \([T^2]\), resulting in \([L]\).Both terms have dimension \([L]\), matching the dimension of \(x\). Thus, equation (b) passes dimensional analysis.

Step 4: Check Equation (c)

Equation (c) is \(x = v_0 t + 2a t^2\). The dimensional analysis is as follows:- Term \(v_0 t\): Dimension of \(v_0\) is \([LT^{-1}]\) and \(t\) is \([T]\), resulting in \([L]\).- Term \(2a t^2\): Dimension of \(a\) is \([LT^{-2}]\) and \(t^2\) is \([T^2]\), resulting in \([L]\).Both terms have dimension \([L]\), thus equation (c) also passes dimensional analysis.

Key Concepts

Kinematics

In physics, understanding the motion of objects is essential. Kinematics is the branch that deals with these motions without considering the causes. It focuses on parameters like distance, speed, velocity, and acceleration. These parameters help in describing how an object moves over time. An essential aspect of kinematics is understanding how to derive equations that can predict future states of motion, given initial conditions. For instance, calculations for distance traveled using speed and time or solving for change in velocity using acceleration are typical problems. Accurate kinematic equations rely on having proper dimensions, ensuring that each term of the equation reflects a comparable physical quantity. This is where dimensional analysis comes into play, ensuring that the calculations and predicted values align with real-world behavior. With the right understanding of kinematic principles, one can predict an object's position, velocity, or acceleration at any future time.

Physics Problem Solving

Physics problem-solving often involves a structured approach, which can appear daunting at first. However, breaking down the process can make it manageable and often simpler than anticipated.

• Identify what you're solving: Determine the physical concepts involved, such as forces, energy, or motion.
• List down known variables: Write out what information is given, including any initial conditions.
• Choose equations wisely: Based on known variables and the required unknowns, select appropriate formulas.
• Verify dimensional consistency: Check that every equation you use produces results in the correct units or dimensions.
• Solve and reflect: Complete the mathematical process to find a solution, then reflect if it makes physical sense.

Taking these steps systematically can ease the process significantly. A solution that aligns perfectly with expected dimensions likely suggests accuracy, providing insights into evolving real-world scenarios.

Equation Verification

Equation verification is a critical part of solving physics problems, especially in kinematics. When you derive or encounter an equation, verifying its correctness is crucial to trust its results. This process involves making sure that the equations match across all terms concerning their units and dimensions. The method of dimensional analysis is essential here. By checking each term of the equation against its expected unit of measure, one can ensure that the equation is dimensionally consistent. For instance, if you have an equation for distance, every single term should resolve to a unit of length [L]. This kind of check helps eliminate errors and misunderstandings before calculations become complex. Both equations (b) and (c), in this problem, passed dimension checks. This means they have potential validity for predicting true physical behavior. A failed dimensional check, as seen with equation (a), signifies something amiss, showing that without dimensional consistency, the results can be misleading.