Proving the Equation s uvt/2 in Kinematics

Understanding and proving the equation for displacement s uvt/2 is crucial for grasping the basics of kinematics, particularly in linear motion with a constant acceleration. This equation, part of the SUVAT equations, is pivotal in solving problems related to motion under uniform acceleration. Let's delve into the proof and explore its application through various methods.

Understanding the SUVAT Equations

The SUVAT equations are a set of five equations that describe the motion of an object moving in a straight line with a constant acceleration. The letters stand for: S: Displacement U: Initial velocity V: Final velocity A: Acceleration T: Time

Proving the Equation s uvt/2

The equation s uvt/2 is derived from the principles of kinematics. Here’s a step-by-step breakdown of the proof:

Method 1: Algebraic Derivation

Let's start with the fundamental equations for motion under constant acceleration:

s ut (1/2)at2 v u at

Note that v u at can be rearranged to find at v - u. Substituting this into the first equation:

s ut (1/2)(v - u)t s ut (1/2)vt - (1/2)ut s (1/2)ut (1/2)vt s (1/2)(ut vt) s (1/2)vt

However, this is not the complete equation. To get the final form, we need to consider the average velocity.

Method 2: Using Average Velocity

The average velocity vavg is defined as the mean of the initial and final velocities:

vavg (u v)/2

The displacement s can then be expressed in terms of average velocity and time:

s vavgt s (u v)/2 * t

This gives us the equation:

s (u v)t/2

Geometric Interpretation Using a Velocity-Time Graph

To further solidify our understanding, let's consider a velocity-time graph where the horizontal axis represents time (t) and the vertical axis represents velocity (v). In a linear motion scenario, the velocity changes at a constant rate from u to v. The area under the velocity-time graph represents the displacement:

Step 1: Draw the velocity-time graph with the vertical axis marked as u and v, and the horizontal axis marked as t. Step 2: The graph is a right triangle with base t and height (v - u).

The area of the triangle can be found using the formula for the area of a triangle:

Area 1/2 * base * height Area 1/2 * t * (v - u)

Rearranging for the displacement:

s v * t - u * t / 2 s (v * t - u * t) * 1/2 s (v * t - u * t) / 2 s (u v) * t / 2

Therefore, we have proven that the equation s uvt/2 is valid under the given conditions.

Applications of s uvt/2

This equation is useful in a variety of scenarios, such as calculating the distance traveled under constant acceleration or verifying the consistency of motion data. For instance, if you have the initial velocity, final velocity, and time, you can easily calculate the displacement.

Example Problem

Consider an example where an object starts with an initial velocity of 2 m/s, reaches a final velocity of 4 m/s after 4 seconds. Using the equation s uvt/2, we can calculate the displacement:

s (2 m/s 4 m/s) * 4 s / 2 s (6 m/s) * 4 s / 2 s 24 m / 2 s 12 m

Thus, the displacement is 12 meters, which can be verified by considering the geometry of the velocity-time graph or by using the SUVAT equation s ut (1/2)at2.

Understanding the derivation and application of the equation s uvt/2 is essential for solving problems in kinematics and solidifies one's grasp of motion under constant acceleration. Always ensure that all the dimensions are compatible and that any unmentioned assumptions are taken into account to avoid errors in your calculations.