Lesson Notes By Weeks and Term v5 - Grade 10

Lesson Video

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Performance objectives

• Define and apply the equations of motion for constant acceleration.
• Solve problems for displacement, velocity, acceleration, and time.
• Analyze motion scenarios to choose the correct equation.
• Differentiate between uniform and non-uniform motion.
• Solve word problems in a South African context.

Lesson summary

This week, we delve deeper into the fascinating world of motion! We're continuing our study of Mechanics, specifically focusing on motion in one dimension. This builds upon the concepts we introduced last week (displacement, velocity, and acceleration) and equips you with the tools to describe and predict how objects move along a straight line. Understanding motion is crucial, not just in physics, but also in understanding everyday phenomena, from driving a taxi in Johannesburg to understanding the trajectory of a cricket ball at the Wanderers stadium. Knowing the physics behind motion allows us to design safer vehicles, analyze sports performance, and even predict weather patterns!

Lesson notes

2.1 Equations of Motion (SUVAT Equations) The equations of motion, often called the SUVAT equations, are a set of equations that relate displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t) for objects moving with constant acceleration in a straight line. It's crucial to remember they ONLY apply when the acceleration is constant.

Here are the four equations: Equation 1: v = u + at (Relates final velocity, initial velocity, acceleration, and time)

Equation 2: s = ut + ½at² (Relates displacement, initial velocity, acceleration, and time)

Equation 3: v² = u² + 2as (Relates final velocity, initial velocity, acceleration, and displacement)

Equation 4: s = ½(u + v)t (Relates displacement, initial velocity, final velocity, and time)

Understanding the Variables: s: Displacement (measured in meters, m). This is the change in position of the object, not necessarily the total distance travelled. u: Initial velocity (measured in meters per second, m/s). This is the velocity of the object at the beginning of the time interval you are considering. v: Final velocity (measured in meters per second, m/s). This is the velocity of the object at the end of the time interval you are considering. a: Acceleration (measured in meters per second squared, m/s²). This is the rate of change of velocity. A positive acceleration means the velocity is increasing (in the positive direction), while a negative acceleration (deceleration or retardation) means the velocity is decreasing. t: Time (measured in seconds, s).

Important Notes: Direction is Crucial: Velocity, displacement, and acceleration are all vector quantities, meaning they have both magnitude and direction. You must choose a positive direction and stick to it consistently throughout the problem. For example, you might choose "up" or "to the right" as positive. If a quantity acts in the opposite direction, it must be given a negative sign.

Constant Acceleration is Key: These equations only work if the acceleration is constant. If the acceleration is changing, you'll need more advanced techniques (which you'll encounter in later studies).

Units: Ensure all quantities are in SI units (meters, seconds, meters per second, meters per second squared) before plugging them into the equations. 2.2 Worked Examples Example 1: A taxi starts from rest at a traffic light in Durban. It accelerates uniformly at a rate of 2.5 m/s² for 6 seconds. a) What is the taxi's final velocity? b) How far does the taxi travel during this time?

Solution: Step 1: Identify the Knowns and Unknowns u = 0 m/s (starts from rest) a = 2.5 m/s² t = 6 s v = ? (unknown, part a) s = ? (unknown, part b)

Step 2: Choose the Appropriate Equation(s) For part (a), we need an equation that relates v, u, a, and t. Equation 1 (v = u + at) is perfect. For part (b), we need an equation that relates s, u, a, and t. Equation 2 (s = ut + ½at²) works.

Step 3: Substitute and Solve (a) v = u + at = 0 + (2.5 m/s²)(6 s) = 15 m/s (b) s = ut + ½at² = (0 m/s)(6 s) + ½(2.5 m/s²)(6 s)² = 0 + (1.25 m/s²)(36 s²) = 45 m Answer: The taxi's final velocity is 15 m/s, and it travels 45 meters.

Example 2: A soccer ball is kicked upwards with an initial velocity of 12 m/s. Assume that the acceleration due to gravity is -9.8 m/s² (negative because it acts downwards). a) How long does it take for the ball to reach its highest point? b) What is the maximum height reached by the ball?

Solution: Step 1: Identify the Knowns and Unknowns u = 12 m/s (positive, we're choosing upwards as positive) a = -9.8 m/s² v = 0 m/s (at the highest point, the ball momentarily stops) t = ? (unknown, part a) s = ? (unknown, part b)

Step 2: Choose the Appropriate Equation(s) For part (a), we need an equation that relates v, u, a, and t. Equation 1 (v = u + at) works. For part (b), we need an equation that relates v, u, a, and s. Equation 3 (v² = u² + 2as) works.

Step 3: Substitute and Solve (a) v = u + at => 0 = 12 m/s + (-9.8 m/s²)t => 9.8t = 12 => t = 12/9.8 = 1.22 s (approximately) (b) v² = u² + 2as => 0² = (12 m/s)² + 2(-9.8 m/s²)s => 0 = 144 - 19.6s => 19.6s = 144 => s = 144/19.6 = 7.35 m (approximately)

Answer: It takes approximately 1.22 seconds for the ball to reach its highest point, and the maximum height reached is approximately 7.35 meters.

Example 3: A bakkie is travelling at a constant velocity of 20 m/s along the N

1. The driver sees a cow in the road and applies the brakes, causing the bakkie to decelerate at a rate of 4 m/s². a) How long does it take for the bakkie to come to a stop? b) What distance does the bakkie travel while braking?

Solution: Step 1: Identify the Knowns and Unknowns u = 20 m/s a = -4 m/s² (negative because it's deceleration) v = 0 m/s (comes to a stop) t = ? (unknown, part a) s = ? (unknown, part b)

Step 2: Choose the Appropriate Equation(s) For part (a), we need an equation that relates v, u, a, and t. Equation 1 (v = u + at) is perfect. For part (b), we need an equation that relates v, u, a, and s.