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Chapter 24: Uniform Circular Motion
Avector is the name of the thing.
A change in direction is a change in the vector, so an acceleration is required to change a velocity's direction.
The direction of the velocity will be changed by the direction of the acceleration.
Uniform circular motion is achieved when the direction is the only quantity changing.
An object is undergoing periodic, uniform circular motion.
We mean that the object maintains a constant speed as it revolves around a circle for a period of time.
The number of revolutions per second is called the Frequency.
The centripetal acceleration is the direction of the acceleration.
It is more convenient to describe the motion in terms of the radians.
The rate of change of the position of the body is also known as the rate of rotation.
The mass is moving with a constant speed of 10 m/s in a circle of 2 m.
All three of the displacement, velocity, and acceleration are related.
The distance, speed, and time are all variables.
There is a description of motion.
The rate of change of displacement is equal to the quotient.
The rate of change of velocity is what Acceleration is defined to be.
The slope of a displacement versus time graph is called the vechicle.
The slope of a graph is called the Acceleration.
The area under the graph can be used to get the displacement.
The acceleration is caused by gravity near the surface of Earth and is directed downward.
This is the only way to get rid of free fall and projectile motion problems.
The relative velocity can be found by adding the individual velocities.
In the absence of air resistance, the horizontal motion is independent of the vertical motion.
The launch velocity can be obtained using the vertical and horizontal components.
The regular equations can be used for each direction.
The centripetal acceleration is when an object moves in a circle.
When you solve a physics problem, be sure to consider the assumptions being made about the moving object.
You will be able to keep track of what is relevant for your solution path in this way.
The goals may be explicit or implicit.
If a question is based on a decision or prediction, you need to understand the requirements to reach an answer.
Consider the meaning of your solution.
Remember the sign conventions for treating vector quantities when choosing a coordinate system.
Make sure you understand the nature of the concepts being discussed.
Make sure that correct units are included in your final answer by using proper SI units throughout your calculations.
Try to figure out if the answer makes sense.
Maybe it looks too large or small because it's expressed in the wrong units.
If no sketch is provided, make one.
If you are interpreting a graph, you need to understand the interrelationships of all the variables.
If you want to make a graph, you need to label both axes, choose a scale for each axis, and draw clearly.
If you get stuck on a difficult problem, try different problem-solving tricks.
If the problem is two-dimensional, break the vectors into components first.
A ball with an initial speed of 20 m/s is thrown upward.
A plane lands on a runway with a speed of 150 m/s.
A ball is thrown from a roof at 25 m/s.
After 25 m/s 2 for 5 s, the engine is shut off and the rocket continues to move upward.
The velocity versus time graph is shown below.
An object has a speed of 15 m/s.
A projectile is launched with a speed of 250 m/s.
A projectile is launched from the top of a 75-m height.
A projectile is launched.
500 m away, it hits the top of a building.
The operator of a boat wants to cross a 5-km wide river that is flowing to the east at 10 m/s.
He wants to reach the exact point on the opposite shore after 15 minutes.
The graph of velocity versus time can be made if the particle begins its motion at t.
A stone is dropped.
A second stone is thrown downward at the same time that the first stone hits the ground.
A particle is moving in one direction.
A stone is dropped from a height and falls in 4 seconds.
A girl standing on top of a roof throws a stone into the air.
She throws a stone with the same speed.
When the stones reach the ground, compare their velocities.
A mass attached to a string is twirled overhead in a horizontal circle.
The mass lands 2.6 m away.
A football quarterback throws a pass to a receiver at an angle of 25 degrees to the horizontal and at an initial speed of 25 m/s.
The quarterback is 30 m from the receiver.
The receiver runs at a constant pace to catch the ball.
A car is moving in a straight line.
The raindrops are falling with a constant terminal velocity.
We need to know how long it will take to decelerate the ball.
The answer is 2.04 s if you divide 20 m/s by 9.8 m/s 2.
The rocket has a speed of 125 m/s after 5 s of acceleration.
The rocket is decelerated by gravity as it continues to move upward after the engine stops.
The time to decelerate to zero is found by dividing 125 m/s by gravity.
The first accelerated distance is added to the distance traveled during that time.
The total distance traveled is equal to 1,100 m.
The areas of the triangles and rectangles add up to 62.5 m.
35 - 2.5 - 20 - 5 is the final displacement.
The change in displacement from question 6 is 7.5 m in 14 s and the average velocity is 0.53 m/s.
If the average speed is 30 m/s, the final speed must be 45 m/s.
A change of 30 m/s at a rate of 3 m/s 2 for 10 s is implied.
We get vx with the known numbers, which is 175 m/s.
We get 3.81 s for the time if we substitute known numbers.
Therefore, vi cos th is 125, vi sin is 32.1, and tan is 0.2568.
The Pythagorean theorem states that the boat must be at least 11 m/s.
The angle is given by the function.
The W of N is 60.9deg.
The graph shows the constant acceleration from t to t. The area has a constant change in speed from 0 to 9 m/s.
The object slows down and turns around in the second region.
The area is -12 m/s.
The graph went from 9 m/s to 3 m/s.
The last area has a change of 3 m/s.
The first stone will be dropped from a height of 75 m, and it will be in free fall because of gravity.
Since the first stone is to fall 15 m before the second, we can determine that t is 1.75 s to fall that distance.
The second stone must reach the ground after the first one has traveled for 3.8 s.
The second stone has a downward initial velocity of 24.34 m/s.
The average velocity is equal to 15 m/s for D x and D t. The new average velocity is equal to 13.5 m/s.
We are continuing this procedure.
The average is 12.6 m/s.
The average is 12.3 m/s.
The average is 12.03 m/s.
The average velocity is 12.003 m/s.
The instantaneous velocity is equal to 12 m/s at 2 s.
T is the total time to fall.
The stone falls in 4 seconds.
h is the number of meters, and T is the number of seconds.
The average velocity is the ratio of the change in displacement to the change in time.
It is possible that the object stops and continues.
It is possible to have a zero instantaneous velocity.
When the stones reach the ground, they will have the same speed.
When the first stone rises, gravity decelerates it until it stops, and then it falls back down.
It has the same speed but in a different direction when it passes its starting point.
The starting speed is the same as the second stone.
Both stones are accelerated through the same displacement, giving them the same final velocities.
The total displacement is divided by the total time to arrive at the average velocity.
The average speed is the same as the total distance divided by the time.
An object can have zero displacement in one period if it returns to its starting point.
It has zero average velocity.
Since it traveled a long way, it has an average speed.
The mass is moving in a straight line.
It has a constant speed.
The height of the mass and the horizontal range can be used to determine thevelocity when it is released.
We find that the r is 0.288 m.
The receiver has to travel 18.85 m away from the quarterback to catch the ball.
To determine how fast the receiver must run, we need to know how long it takes the ball to travel.
To travel 18.85 m in 2.16 s, the receiver needs to run at 8.73 m/s.
The driver can see that from the diagram below.
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