Physics

One-Dimensional Kinematics

One-dimensional kinematics is the study of motion in a straight line. It involves analyzing how an object's position, velocity, and acceleration change over time.

• Position: The position of an object is its location in space. It is often described in terms of an object's distance from a reference point.

• Velocity: The velocity of an object is its speed in a specific direction. It is a vector quantity, meaning it has both a magnitude (speed) and a direction.

• Acceleration: The acceleration of an object is the rate at which its velocity changes. It is also a vector quantity, with both a magnitude and a direction.

• To analyze one-dimensional motion, we use equations of motion. These equations relate an object's position, velocity, and acceleration at different times. The most commonly used equations of motion are:

  • Position equation: x = x0 + v0t + (1/2)at^2 This equation describes how an object's position changes over time, given its initial position (x0), initial velocity (v0), acceleration (a), and time (t).

  • Velocity equation: v = v0 + at This equation describes how an object's velocity changes over time, given its initial velocity (v0), acceleration (a), and time (t).

  • Acceleration equation: a = (v-v0)/t This equation describes how an object's acceleration changes over time, given its initial velocity (v0), final velocity (v), and time (t).

• Frame of reference: A frame of reference is a set of points or objects that are used to describe the position of other objects. It is important to specify a frame of reference when analyzing motion, because the way we perceive an object's motion can depend on our own motion relative to the object. For example, an airplane appears to be moving slower to a person on the ground than it does to a person on board the plane.

Displacement and distance:

• Displacement is the change in an object's position, or the distance between an object's starting position and its ending position.

• Distance is the total length of the path traveled by an object, regardless of direction.

  • For example, if an object travels in a straight line from point A to point B, its displacement is the distance between A and B, but its distance traveled is the length of the path it followed (which could be longer than the displacement if the object traveled in a zigzag pattern).

Velocity and speed:

• Velocity is an object's speed in a specific direction.

• Speed is an object's rate of motion, but it does not take direction into account.

• For example, if an object moves at a constant speed of 50 km/hr to the east, its velocity is 50 km/hr to the east. If the same object then turns around and travels 50 km/hr to the west, its velocity is now -50 km/hr to the west (the negative sign indicates the opposite direction). Its speed, however, remains constant at 50 km/hr in both cases.

Average velocity:

• The equation for average velocity can be used to find the average speed of an object over a certain time period. It is given by the formula: average velocity = (final position - initial position) / time elapsed. This equation is most useful when an object's velocity is changing over time, as it gives a single value that represents the overall motion of the object.

Meaning of slope and area on displacement, velocity, and acceleration graphs:

• The slope of a graph is a measure of its steepness.

• On a displacement-time graph, the slope represents an object's velocity (a steep slope indicates a high velocity, while a shallow slope indicates a low velocity). On a velocity-time graph, the slope represents an object's acceleration (a steep slope indicates a high acceleration, while a shallow slope indicates a low acceleration).

• The area under a graph can also provide information about an object's motion.

  • On a displacement-time graph, the area under the graph represents the distance traveled by the object. On a velocity-time graph, the area under the graph represents the change in an object's position (displacement).

4 Kinematic Equations:

The four kinematic equations (position, velocity, acceleration, and acceleration) can be used to solve for unknown motion variables, given the values of other variables. For example, if you know an object's initial position, final position, and time elapsed, you can use the position equation to find its initial velocity. Or, if you know an object's initial velocity, acceleration, and time elapsed, you can use the velocity equation to find its final velocity.

Two-Dimensional Kinematics

Two-dimensional kinematics is the study of motion in two dimensions, such as on a plane or a flat surface. It involves analyzing how an object's position, velocity, and acceleration change over time in two dimensions.

• In two-dimensional kinematics, position is often described in terms of an object's coordinates, which are pairs of numerical values that specify the object's location in a plane. For example, an object's position might be described as (x, y), where x and y are the object's distances from the x-axis and y-axis, respectively.

• Velocity in two dimensions is a vector quantity, meaning it has both a magnitude (speed) and a direction.

• Acceleration in two dimensions is also a vector quantity, with both a magnitude and a direction.

• To analyze two-dimensional motion, we use the same equations of motion that are used in one-dimensional kinematics, but we must consider the x and y components of the object's position, velocity, and acceleration separately. The equations of motion for two-dimensional kinematics are:

  • Position equations: x = x0 + v0xt + (1/2)axt^2 and y = y0 + v0yt + (1/2)ayt^2 These equations describe how an object's position in the x and y dimensions changes over time, given its initial position (x0, y0), initial velocity (v0x, v0y), acceleration (ax, ay), and time (t).

  • Velocity equations: vx = v0x + axt and vy = v0y + ayt These equations describe how an object's velocity in the x and y dimensions changes over time, given its initial velocity (v0x, v0y), acceleration (ax, ay), and time (t).

  • Acceleration equations: ax = (vx-v0x)/t and ay = (vy-v0y)/t These equations describe how an object's acceleration in the x and y dimensions changes over time, given its initial velocity (v0x, v0y), final velocity (vx, vy), and time (t).

Vectors:

• Calculating the component of a vector: To calculate the component of a vector in a particular direction, you can use the scalar projection formula. This formula states that the component of a vector (v) in a certain direction (d) is equal to the dot product of the vector and the unit vector in the direction of interest, divided by the magnitude of the unit vector:

  • component = (v * d) / |d|

    • Where * represents the dot product, and |d| represents the magnitude of d.

Finding the resultant of two vectors: The resultant of two vectors is the vector that represents their combined effect. There are several ways to find the resultant of two vectors, including:

• Graphical method: You can find the resultant of two vectors graphically by drawing the vectors on a graph and then drawing a vector from the starting point of the first vector to the ending point of the second vector. The resultant vector is the vector that connects the starting point to the ending point.

• Analytical method: You can use simple trigonometry to find this as vectors should make a right triangle

Projectile Motion:

• Describing the two-dimensional motion of a projectile: A projectile is an object that is thrown, shot, or otherwise projected into the air. The motion of a projectile is affected by the force of gravity and the initial velocity at which it is launched. To describe the two-dimensional motion of a projectile, you can use the equations of motion for two-dimensional kinematics, taking into account the effects of gravity. For example, the position of a projectile at any given time (t) can be calculated using the following equations:

• x = x0 + v0xt + (1/2)axt^2 y = y0 + v0yt - (1/2)gt^2

• Where x0 and y0 are the projectile's initial position, v0x and v0y are its initial velocity in the x and y dimensions, ax and ay are its acceleration in the x and y dimensions (which are zero for a projectile in the absence of other forces), and g is the acceleration due to gravity (which is negative in the downward direction).

Uniform Circular Motion:

• Uniform circular motion is the motion of an object moving in a circular path at a constant speed. It is characterized by a constant centripetal acceleration, which is the acceleration that is directed towards the center of the circular path and is necessary to keep the object moving in a circular path. The magnitude of the centripetal acceleration is given by the formula:

  • a = v^2 / r

    • Where v is the object's speed and r is the radius of the circular path.

• In uniform circular motion, the direction of the centripetal acceleration is always towards the center of the circular path. This means that the object is constantly changing direction, even though it is moving at a constant speed. The change in direction is caused by the object's velocity vector constantly changing direction, while its magnitude remains constant.

• There are several applications of uniform circular motion in everyday life, including the motion of a car turning a corner and the motion of a satellite orbiting a planet.

• Centripetal acceleration: a = v^2 / r This formula gives the magnitude of the centripetal acceleration, which is the acceleration that is directed towards the center of the circular path.

• Centripetal force: F = mv^2 / r This formula gives the magnitude of the centripetal force, which is the force that is necessary to produce the centripetal acceleration.

• Angular velocity: ω = v / r This formula gives the angular velocity of an object in uniform circular motion, which is the speed at which the object rotates around the center of the circular path.

• Angular displacement: θ = ωt This formula gives the angular displacement of an object in uniform circular motion, which is the change in the object's angular position over time.

Differences between uniform circular motion and projectile motion: Uniform circular motion is motion in a circular path at a constant speed. Projectile motion is motion under the influence of gravity, with an initial velocity in a specific direction. Some key differences between the two types of motion include:

• Path of motion: In uniform circular motion, an object moves in a circular path at a constant speed. In projectile motion, an object follows a curved path due to the influence of gravity.

• Acceleration: In uniform circular motion, an object experiences a constant acceleration (called centripetal acceleration) towards the center of the circular path. In projectile motion, an object experiences a constant acceleration (due to gravity) in the downward direction.

• Forces: In uniform circular motion, the only force acting on an object is the centripetal force, which is directed towards the center of the circular path. In projectile motion, there are two forces acting on an object: gravity and air resistance (which is typically negligible).

Newton's Laws

• Newton's first law: Newton's first law, also known as the law of inertia, states that a body at rest tends to stay at rest, and a body in motion tends to stay in motion with a constant velocity, unless acted upon by a force. This law is based on the idea that an object will stay at rest or continue moving in a straight line at a constant speed unless something forces it to change.

  1. Inertia is the property of an object that resists a change in its state of motion. An object with more mass has more inertia, meaning it is harder to change its motion. For example, it is easier to push a light object across a table than it is to push a heavy object, because the light object has less inertia.

• Newton's second law: Newton's second law states that the force applied to a body is equal to the mass of the body multiplied by its acceleration. This law is often written as F = ma, where F is the force, m is the mass of the body, and a is the acceleration. It states that the more massive an object is, the more force is required to accelerate it, and the more quickly an object is accelerated, the more force is required. Net force is the total force acting on an object.

  1. When an object is subject to multiple forces, the net force is the vector sum of all the forces. If the net force on an object is not zero, the object will accelerate in the direction of the net force, with an acceleration that is proportional to the net force and inversely proportional to the mass of the object. For example, if you apply a force of 10 N to a 1-kilogram object, it will accelerate at a rate of 10 m/s^2. If you apply the same force to a 2-kilogram object, it will accelerate at a rate of 5 m/s^2.

  2. The main equation is F=ma (force = mass*acceleration)

• Free Body Diagram (FBD): A Free Body Diagram (FBD) is a graphical representation of an object that shows all the forces acting on it. An FBD is used to help visualize the forces and understand how they are affecting the object. To draw an FBD, you start by drawing a box or other simple shape to represent the object. Then, you draw arrows to represent the forces acting on the object, with the length of the arrow representing the magnitude of the force and the direction of the arrow representing the direction of the force.

• Newton's third law: Newton's third law states that for every action, there is an equal and opposite reaction. This law states that whenever one body exerts a force on another body, the second body exerts an equal and opposite force back on the first body. For example, if you push a book across a table, the book will push back on your hand with an equal and opposite force. This law is based on the idea that forces always occur in pairs, and that the two forces in a pair are equal in magnitude and opposite in direction.

Practice Problems:

A big part of physics is being able to know what formula to use. In 1D Kinematics this is not too dificult but becomes harder as physics goes on.

• An object is initially located at x0 = 3 meters and has an initial velocity of v0 = 5 m/s. It accelerates at a rate of a = 2 m/s^2 for t = 4 seconds. Find the object's final position.

Solution: x = 3 + (5)(4) + (1/2)(2)(4^2) = 3 + 20 + 16 = 39 meters

• An object has an initial velocity of v0 = 10 m/s and accelerates at a rate of a = 3 m/s^2 for t = 5 seconds. Find the object's final velocity.

Solution: v = 10 + (3)(5) = 10 + 15 = 25 m/s

• An object has an initial velocity of v0 = 20 m/s and a final velocity of v = 30 m/s after t = 2 seconds. Find the object's acceleration.

Solution: a = (30 - 20)/2 = 10/2 = 5 m/s^2

• An object starts at x0 = 5 meters and ends at x = 10 meters after t = 3 seconds. Find the object's average velocity.

Solution: average velocity = (10 - 5)/3 = 5/3 = 1.67 m/s

Practice Problems:

Problem 1:

A projectile is launched at an angle of θ = 30° and an initial velocity of v0 = 50 m/s. The projectile lands on a flat surface 3 seconds later. Calculate the projectile's position (x and y coordinates) at the time of landing.

Solution:

First, we need to find the x and y components of the projectile's initial velocity: v0x = v0cosθ = 50cos30° = 43.301 m/s v0y = v0sinθ = 50sin30° = 25 m/s

Next, we can use the equations of motion to find the projectile's position at the time of landing: x = x0 + v0xt + (1/2)axt^2 = 0 + (43.301)(3) + (1/2)(0)(3^2) = 129.903 m y = y0 + v0yt - (1/2)gt^2 = 0 + (25)(3) - (1/2)(-9.81)(3^2) = 46.5 m

So, the projectile's position at the time of landing is (129.903, 46.5).

Problem 2:

A projectile is launched at an angle of θ = 45° and an initial velocity of v0 = 50 m/s. The projectile lands on a flat surface 4 seconds later. Calculate the projectile's velocity (magnitude and direction) at the time of landing.

Solution:

First, we need to find the x and y components of the projectile's initial velocity: v0x = v0cosθ = 50cos45° = 35.355 m/s v0y = v0sinθ = 50sin45° = 35.355 m/s

Next, we can use the equations of motion to find the projectile's velocity in the x and y dimensions at the time of landing: vx = v0x + axt = (35.355) + (0)(4) = 35.355 m/s vy = v0y - gt = (35.355) - (-9.81)(4) = 67.755 m/s

To find the projectile's velocity (magnitude and direction) at the time of landing, we can use these values to convert to polar coordinates: v = √(vx^2 + vy^2) = √(35.355^2 + 67.755^2) = 75.102 m/s θ = atan(vy/vx) = atan(67.755/35.355) = 63.435°

So, the projectile's velocity at the time of landing is 75.102 m/s at an angle of 63.435°.

Practice Problems:

Problem 1:

An object with a mass of 5 kilograms is being pushed across a surface by a force of 10 N. Calculate the velocity of the object after 5 seconds if the object starts from rest.

Solution:

First, we need to find the acceleration of the object: a = F / m = 10 N / 5 kg = 2 m/s^2

Next, we can use the equation v = v0 + at to find the velocity of the object: v = 0 + (2 m/s^2)(5 s) = 10 m/s

So, the velocity of the object after 5 seconds is 10 m/s.

Problem 2:

A force of 15 N is acting on an object with an acceleration of 3 m/s^2. Calculate the mass of the object.

Solution:

According to Newton's second law (F = ma), the mass of the object is given by: m = F / a = 15 N / 3 m/s^2 = 5 kg

So, the mass of the object is 5 kg.Problem 3:

A person is pushing a box across a floor. Identify the action force and the reaction force in this situation.

Solution:

The action force in this situation is the force that the person is applying to the box, which is the force that is causing the box to accelerate. The reaction force in this situation is the force that the box is applying to the person, which is equal in magnitude and opposite in direction to the force that the person is applying to the box.

So, the action force in this situation is the force applied by the person to the box, and the reaction force is the force applied by the box to the person.