Problem 37 - Physics for Scientists and Engineers with Modern Physics

Question

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

Step-by-Step Solution

Answer

Equations (b) and (c) are dimensionally correct.

Step 1: Analyze the Equation (a)

The equation is given by \( x = vt^2 + 2at \). First, check each term for dimensional consistency. The left-hand side (LHS) is distance \([x] = L\). For the right-hand side (RHS), the term \( vt^2 \) has dimensions \([v] = L/T\) and \([t^2] = T^2\), giving \( L \cdot T \), which is incorrect because it doesn’t reduce to \( L \). The term \( 2at \) has dimensions \([a] = L/T^2\) and \([t] = T\), resulting in \( L \). The second term is dimensionally consistent, but the first term is not, so equation (a) is dimensionally incorrect.

Step 2: Analyze the Equation (b)

The equation is given by \( x = v_0 t + \frac{1}{2} at^2 \). Checking each term: LHS \([x] = L\); for RHS, the term \( v_0 t \) has dimensions of speed \([v_0] = L/T\) and time \([t] = T\) resulting in \( L \), which is correct. For the second term \( \frac{1}{2} at^2 \), the dimensions of acceleration \([a] = L/T^2\) and \([t^2] = T^2\) results in \( L \). Both terms are dimensionally consistent, so equation (b) could be correct.

Step 3: Analyze the Equation (c)

The equation is given by \( x = v_0 t + 2at^2 \). Checking each term: LHS \([x] = L\); for RHS, the first term \( v_0 t \) is already verified as \( L \) from previous analysis in equation (b), which is consistent. For the second term \( 2at^2 \), dimensions are \([a] = L/T^2\) and \([t^2] = T^2\), which gives \( L \). Both terms are dimensionally consistent, so equation (c) could also be correct.

Key Concepts

Dimensional Consistency

Dimensional consistency is a fundamental principle in physics that insists equations must be dimensionally coherent. It ensures that the dimensions on both sides of an equation match, thus making the equation physically meaningful. To check for dimensional consistency, we assess each term individually in terms of its basic dimensions:

• Length \([L]\)
• Time \([T]\)
• Mass \([M]\)

In the original exercise, we checked each equation by breaking it down into its component parts and examining the dimensions:

• The distance \([x] = L\) on the left-hand side must be matched by equivalent dimensions on the right-hand side.
• For example, in the equation \(x = vt^2 + 2at\), the term \(vt^2\) has dimensions \([v] = L/T\) and \([t^2] = T^2\), which combined gives \(L \cdot T\), not simplifying to \([L]\). This shows it's incorrect because it breaks dimensional consistency.

Understanding and applying dimensional consistency in equations is crucial because it helps verify whether an equation could conceptually be correct before even diving into numerical calculations.

Kinematic Equations

Kinematic equations are widely used in physics to describe the motion of objects. They relate the quantities of velocity, acceleration, displacement, and time. Typically these equations, such as

• \( v = u + at \)
• \( x = ut + \frac{1}{2} a t^2 \)
• \( v^2 = u^2 + 2ax \)

are used when an object moves in a straight line with uniform acceleration.
In the exercise, the given equations aspire to describe such motion. However, only some equations passed the dimensional consistency test to be potential candidates for accurate kinematic equations:

• \(x = v_0 t + \frac{1}{2} at^2\), for example, could be a legitimate kinematic equation given that its terms introduce displacement consistently with dimension \([L]\).
• This affirms that understanding how to derive or check these equations is critical, as they form the backbone of solving problems in dynamics.

Distance and Motion Analysis

Distance and motion analysis in physics involves determining how objects move and the distance they cover over time. This falls under the broader category of kinematics, which aims to detail the trajectory of an object without considering the forces causing movement.
The fundamental approach involves

• Defining the initial conditions, such as initial velocity (\(v_0\)
• Acceleration (\(a\)

Evaluating the progress using given time (\(t\).
Throughout the exercise, each equation attempted to model distance traveled under specific conditions:

• For example, equations with terms like \(v_0 t\) and \(\frac{1}{2} at^2\), which accurately reflected the initial velocity's contribution and the effect of constant acceleration over time.
• Such components are crucial for precisely calculating how far and how fast an object travels under uniform conditions.

Accurate interpretation of these elements is central for problem-solving in real-world physics applications, providing a deeper understanding of motion analysis.