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CIE A-Level Physics Notes

2.1.3 Acceleration

Exploring Acceleration

Acceleration, as a vector quantity, encompasses both the rate of change in speed and the direction of motion.

  • Definition: Acceleration is the rate of change of velocity per unit time.
  • Formula and Units: Acceleration (a) = Change in Velocity (Δv) / Change in Time (Δt), measured in meters per second squared (m/s²).
Diagram explaining acceleration

Acceleration

Image Courtesy GeeksforGeeks

Types of Acceleration

Acceleration is categorised based on how it affects an object's motion.

1. Positive Acceleration: When an object's velocity increases over time. If an object moves in a negative direction, positive acceleration means it's getting faster in that direction.

  • Example: A car speeding up from 20 m/s to 50 m/s in 5 seconds.

2. Negative Acceleration (Deceleration): When an object's velocity decreases. In a negative direction, it implies slowing down in that direction.

  • Example: A bus slowing down from 30 m/s to 10 m/s over 4 seconds.

3. Zero Acceleration: When an object maintains a constant velocity, indicating no change in speed or direction.

  • Example: A cruise ship travelling at a steady speed of 30 m/s.
Diagram explaining the comparison between positive acceleration and negative acceleration

Positive and negative acceleration

Image Courtesy Khan Academy

Detailed Sample Calculations

Scenario 1: Positive Acceleration

  • Situation: A motorbike increases its speed from 0 to 30 m/s in 6 seconds.
    • Change in Velocity (Δv) = 30 m/s - 0 = 30 m/s
    • Change in Time (Δt) = 6 seconds
    • Acceleration (a) = 30 m/s / 6 s = 5 m/s²

Scenario 2: Negative Acceleration

  • Situation: A cyclist going at 10 m/s comes to a stop in 5 seconds.
    • Change in Velocity (Δv) = 0 - 10 m/s = -10 m/s
    • Change in Time (Δt) = 5 seconds
    • Acceleration (a) = -10 m/s / 5 s = -2 m/s²

Scenario 3: Zero Acceleration

  • Situation: A hoverboard moves at a steady speed of 8 m/s for 10 seconds.
    • Δv = 0 m/s (since the speed is constant)
    • Acceleration (a) = 0 m/s / 10 s = 0 m/s²

Practical Applications of Acceleration

Real-Life Examples

  1. Vehicles: Acceleration is experienced when a vehicle speeds up or slows down.
  2. Athletics: Runners accelerate from the start line and decelerate as they finish.

Graphical Representation

  • Velocity-Time Graphs: The slope of a velocity-time graph indicates acceleration. A positive slope denotes positive acceleration, a negative slope denotes negative acceleration, and a flat line denotes zero acceleration.
Diagram explaining the velocity time graph when the acceleration is constant

Velocity-time graphs

Image Courtesy BYJU’s

Acceleration in Physics Principles

Acceleration is a key component in dynamics and kinematics, and it's critical in Newton's laws of motion.

Using Acceleration to Calculate Displacement

The displacement of an object starting from rest and moving with constant acceleration is given by s = 1/2 * a * t2, where s is displacement, a is acceleration, and t is time.

Advanced Aspects of Acceleration

Complex Motion Scenarios

  1. Direction Change: Acceleration occurs with changes in direction, even if speed is constant.
  2. Projectile Motion: Objects in projectile motion experience constant gravitational acceleration, affecting vertical velocity.

Acceleration in Varied Motions

In complex motions, like in aerodynamics, acceleration can vary with velocity and time, influenced by factors like air resistance.

FAQ

Instantaneous acceleration refers to the acceleration of an object at a specific point in time. It is the rate at which an object's velocity changes at a particular instant. This concept is crucial in scenarios where acceleration is changing, such as a car navigating through traffic. On the other hand, average acceleration is defined over a period of time. It is calculated as the change in velocity divided by the time taken for this change. While instantaneous acceleration provides a snapshot of an object’s motion at a particular moment, average acceleration gives an overview of its motion over a time interval. The distinction is especially important in understanding complex motions where acceleration is not constant.

An object can indeed have zero acceleration while still moving. This scenario occurs when the object is moving at constant velocity. Constant velocity means that both the speed and the direction of the object’s motion are unchanging. In this situation, since acceleration is defined as the rate of change of velocity, and there is no change in velocity, the acceleration is zero. An everyday example of this would be a car cruising at a steady speed on a straight road. Despite its motion, if the speed and direction remain constant, the car’s acceleration is zero.

In circular motion, acceleration plays a critical role, known as centripetal acceleration. This acceleration is always directed towards the center of the circle around which the object is moving. It is important to note that even if the speed of the object remains constant, there is still acceleration because the direction of the object's velocity is continually changing. Centripetal acceleration is responsible for changing the direction of the velocity vector, not its magnitude. The formula for centripetal acceleration is a = v² / r, where v is the velocity of the object and r is the radius of the circle. This type of acceleration ensures that the object follows a circular path rather than moving off in a straight line according to Newton's First Law of Motion.

Air resistance significantly impacts the acceleration of falling objects, especially at higher velocities. When an object falls, it accelerates due to gravity, initially following the equation a = g (where g is the acceleration due to gravity, approximately 9.8 m/s²). However, as the object speeds up, air resistance increases, acting in the opposite direction to the motion. This resistance reduces the net acceleration of the object. Eventually, the object reaches a point where the force due to air resistance equals the gravitational force, resulting in zero net force and hence zero acceleration. At this point, the object continues to fall at a constant velocity, known as the terminal velocity. The effect of air resistance is particularly noticeable in objects with large surface areas or those falling from great heights.

Acceleration is intricately related to Newton's Second Law of Motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma). This law implies that acceleration is directly proportional to the force applied and inversely proportional to the mass of the object. For instance, if you apply a greater force to an object, its acceleration increases. Conversely, if the same force is applied to an object with a larger mass, the acceleration decreases. Newton's Second Law essentially provides a quantitative description of how forces affect the motion of objects, with acceleration being a key component of this relationship.

Practice Questions

A car accelerates uniformly from rest to a speed of 60 m/s in 10 seconds. Calculate the acceleration and the distance travelled by the car during this time.

The acceleration can be calculated using the formula a = Δv / Δt. Here, the change in velocity (Δv) is 60 m/s (since it starts from rest), and the change in time (Δt) is 10 seconds. Thus, the acceleration (a) is 60 m/s / 10 s = 6 m/s². To find the distance travelled, we use the formula s = 1/2 * a * t2. Substituting the values, s = 1/2 * 6 m/s² * (10 s)² = 300 meters. Therefore, the car accelerates at 6 m/s², and the distance travelled in this time is 300 meters.

The acceleration can be calculated using the formula a = Δv / Δt. Here, the change in velocity (Δv) is 60 m/s (since it starts from rest), and the change in time (Δt) is 10 seconds. Thus, the acceleration (a) is 60 m/s / 10 s = 6 m/s². To find the distance travelled, we use the formula s = 1/2 * a * t^2. Substituting the values, s = 1/2 * 6 m/s² * (10 s)² = 300 meters. Therefore, the car accelerates at 6 m/s², and the distance travelled in this time is 300 meters.

At the highest point, the velocity of the ball will be 0 m/s. Using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration (which will be -9.8 m/s² as gravity acts downwards), and t is the time taken. Rearranging the formula, t = (v - u) / a. Substituting the values, t = (0 - 20 m/s) / -9.8 m/s² = 2.04 seconds (approximately). Therefore, it takes about 2.04 seconds for the ball to reach its highest point.

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