Lesson 2.2: Equations of Uniformly Accelerated Motion

Introduction

Welcome, students! Today, we are diving into the fascinating world of uniformly accelerated motion, a key concept in Newtonian mechanics. Our objectives for this lesson are:

• Deriving and applying the four "suvat" equations of motion.
• Understanding free fall and the acceleration due to gravity, denoted as $g$.
• Measuring $g$ experimentally.
• Learning about sign conventions for vector quantities in one dimension.
• Solving multi-stage motion problems by combining phases.
• Selecting and applying the appropriate equation of motion to one-dimensional problems.

To get us started, let’s consider a real-world example: when a car accelerates down a straight road or when you drop a ball from a building to see how far it falls. Both scenarios involve uniform acceleration, and we can calculate distances, speeds, and times using specific equations. 🚗💨

The Four Suvat Equations

The equations that govern uniformly accelerated motion are commonly referred to as the "suvat" equations. The term "suvat" comes from the key variables in these equations:

• $s$ = displacement
• $u$ = initial velocity
• $v$ = final velocity
• $a$ = acceleration
• $t$ = time

These equations assume a constant acceleration and are expressed as follows:

1. $$s = ut + \frac{1}{2}at^2$$
2. $$v = u + at$$
3. $$v^2 = u^2 + 2as$$
4. $$s = \frac{(u + v)}{2}t$$

Applying the Suvat Equations

Let’s break down how to use these equations with a practical example:

Example 1: Suppose a car starts from rest (initial velocity $u = 0$ m/s) and accelerates at a rate of $2 \, \text{m/s}^2$ for $5$ seconds. We want to find:

1. The final velocity ($v$) after $5$ seconds.
2. The total displacement ($s$) during this time.

Using the second suvat equation for final velocity:

$$v = u + at = 0 + (2 \, \text{m/s}^2)(5 \, \text{s}) = 10 \, \text{m/s}$$

Now, to find total displacement using the first suvat equation:

$$s = ut + \frac{1}{2}at^2 = (0)(5) + \frac{1}{2}(2)(5^2) = 0 + 25 = 25 \, \text{m}$$

This means after $5$ seconds, the car has traveled $25$ meters and reached a final speed of $10$ m/s! 🎉

Free Fall and Gravity

In our world, one of the most intriguing motions is free fall, where objects fall solely under the influence of gravity. The acceleration due to gravity is generally represented as $g = 9.81 \, \text{m/s}^2$.

Measuring Gravity Experimentally

To measure $g$, you can perform a straightforward experiment by dropping an object from a known height and measuring the time it takes to hit the ground. Based on this data, you can apply the suvat equations to calculate $g$. Here’s a step-by-step process:

1. Drop a ball from a height of $h$ meters.
2. Use a stopwatch to measure how long ($t$ seconds) it takes for the ball to hit the ground.
3. Use the first suvat equation, rearranged to find $g$:

$$g = \frac{2h}{t^2}$$

This simple experiment demonstrates how we can measure one of the most constant forces in our universe! 🌍

Sign Conventions for Vector Quantities

When dealing with motion in one dimension, it's essential to establish a sign convention to avoid confusion.

• Typically, we consider motion in the upward direction as positive and downward motion as negative.
• For example, if an object moves up $5$ m, displacement is $+5 \, \text{m}$; if it falls down $5$ m, displacement is $-5 \, \text{m}$.

Using consistent sign conventions helps simplify calculations and clarifies the direction of motion.

Multi-Stage Motion Problems

In many real-world situations, motion can be broken down into separate phases or stages. For instance, consider a scenario where a car accelerates for $3$ seconds, then travels at a constant speed for $2$ seconds, and finally decelerates for another $3$ seconds.

To solve multi-stage problems, we:

1. Analyze each stage separately using the appropriate suvat equations.
2. Use the final velocity from one stage as the initial velocity for the next.
3. Carefully combine the displacements from each stage to find the total displacement.

Example 2: Multi-Stage Problem

Imagine a car accelerates from rest ($u = 0\, \text{m/s}$) at $2\, \text{m/s}^2$ and it accelerates for $3$ seconds. It then continues at its final speed for $2$ seconds before decelerating at $-1\, \text{m/s}^2$ for another $3$ seconds. What is the total distance traveled?

1. Stage 1:

• Final velocity after acceleration:

$$v = u + at = 0 + (2)(3) = 6 \, \text{m/s}$$

• Displacement during acceleration:

$$s_1 = ut + \frac{1}{2}at^2 = (0)(3) + \frac{1}{2}(2)(3^2) = 0 + 9 = 9 \, \text{m}$$

1. Stage 2:

• Distance at constant speed for $2$ seconds:

$$s_2 = vt = 6(2) = 12 \, \text{m}$$

1. Stage 3:

• Initial velocity here is $v = 6 \, \text{m/s}$, and we are decelerating (\$a = -1\, $\text{m/s}^2$\$):
• Final velocity:

$$v_f = 6 + (-1)(3) = 3 \, \text{m/s}$$

• Displacement:

$$s_3 = (6)(3) + \frac{1}{2}(-1)(3^2) = 18 - 4.5 = 13.5 \, \text{m}$$

1. Total Distance:

$$\text{Total Distance} = s_1 + s_2 + s_3 = 9 + 12 + 13.5 = 34.5 \, \text{m}$$

Conclusion

In this lesson, we explored the equations of uniformly accelerated motion, applied them to real-world scenarios, and understood the concept of free fall and gravity. We also discussed multi-stage problems and the importance of sign conventions in physics.

Study Notes

• The four suvat equations of motion are fundamental to uniformly accelerated motion.
• The formulae are useful for calculating displacement, velocity, and time.
• Free fall involves an acceleration of $g = 9.81 \, \text{m/s}^2$.
• Be consistent with sign conventions for direction in motion.
• Multi-stage problems require analyzing each phase and combining results for total distance.