Showing posts with label Acceleration. Show all posts
Showing posts with label Acceleration. Show all posts

Equations of Motion for Vertically Thrown up body

Equations of motion for a vertically thrown up body

Thus the velocity of the body keeps on decreasing and at a particular height, its velocity becomes zero and that height is said to be the maximum height that the body can go and the time taken to reach that maximum height is called time of ascent. 

Acceleration in this case is acceleration due to gravity which is constant and it shall be treated as negative as the velocity of the body keeps on decreasing with respect to time. Taking this into consideration, we can rewrite the four equations of motion for a freely falling body. It is proved that the velocity with you project the body vertically up is the velocity with which it comes back to the same point of projection but in the opposite direction. It is also proved that when air resistance is ignored, time of ascent is equal to the time of descent.




 Time of Ascent and Decent with Air resistance

As it is mentioned earlier time of ascent of a vertically thrown up body is equal to the time of descent of a freely falling body. The above mentioned statement is true only when air resistance is ignored. But in real life air resistance is there and if that is taken into consideration, time of ascent is different from time of descent. Air resistance offers a force against the motion and it always acts opposite the motion. Thus when the body is thrown up, both gravitational force and resistance force acts against the motion. When the body is coming down, gravitational force acts in down ward direction but air resistance acts opposite to the motion and that is in upward direction. Thus we can find the effective acceleration in both the cases and find the the relations between time of ascent and descent as shown in the video below.




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Motion in a Straight line Introduction and Average Velocity Video Lesson

Motion in a straight line an introduction

Studying the motion of the body without bothering about the forces acting on it is done in kinematics. We treat body as a combination of identical point sized objects and they have negligible dimensions. All laws of mechanics were in principle discussed with the point sized particles and as the body is the combination of similar particles, under ideal conditions the laws are applicable to bodies also. Here we are dealing with bodies moving with a velocity much lesser than the velocity of the light. In this particular case, body is moving only along one dimension either along X,Y or Z axis. This is called one dimensional motion and it is changing its position with respect to time and surroundings.

To measure the change of the position, we have terms like distance, speed. Distance is the actual path traveled by a body and the speed is the rate of change of distance with respect to time. Both distance and speed are treated as scalars and they can be understood by stating their magnitude alone and they don’t need direction.

Displacement is the shortest distance between initial and final positions in specified direction and it is treated as vector quantity. They can be understood completely only when both magnitude and directions are given to us. Velocity is defined as the rate of change of displacement and it is also a vector quantity.




 Average velocity

If a particle is not changing its velocity with respect to time, then it is said to be in uniform velocity. In this case at any given interval of time, the particle will have same constant velocity and it is same every where. But it is not same every where. If a body is changing its velocity with respect to time, then it is having acceleration and we would like to measure the average velocity in the given case. Average velocity is defined as the ratio of total displacement covered by a body in the total time. Taking this concept into consideration, we can find average velocity when time is shared and displacement is shared as shown in the video below.




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Mechanical Properties of Fluids Problems and Solutions Four

We are solving series of problems on fluid dynamics. Viscous force acts on all bodies that have relative motion. If a spherical ball is falling through a fluid, it experience a viscous force and it can be measured using a law called Stroke’s law. According to this rule, viscous force is directly proportional to the square of the radius of the ball. There is weight of the ball acting in the downward direction and upthrust acting always in the upward direction. Viscous force always acts opposite to the relative motion of the body. After some travel, the resultant force acting on the body becomes zero and the body acquires a constant velocity and that velocity is called as terminal velocity. 

Problem

An open U tube contains mercury. When certain amount of water of known height is filled in one tube over mercury, we would like to measure the difference between the liquid levels in both the sides and the problem is as shown in the diagram below.


Solution

We know that at the bottom of the system both the pipes has one common point whose height is same and hence the pressure is same at both the points. We know that the pressure can be expressed as the product of height of the liquid, density of the liquid and acceleration due to gravity at the given place. Thus by equating the pressure at both the points we can solve the problem as shown in the diagram below.


Problem

A sphere of known radius is having a cavity of half the radius of the sphere. It is found that the sphere is just floating in the water with its highest point in touch with the water. We need to find the specific gravity of the material and the problem is as shown in the diagram below.


Solution

We know that when the sphere is just floating, its weight is balanced by the upthrust acting on it. We know that upthrust is the product of volume of the fluid displaced, density of the fluid and acceleration due to gravity. As the body is completely immersed in the fluid, volume of the fluid displaced is equal to the volume of the body itself. We know the values of density of water and acceleration due to gravity.

Weight acting in the downward direction is only due to the mass present in the system. We can write mass as volume and density product of the body. Volume of the content means we shall count only the part of mass present in the system. Thus we can measure the weight and equate it ot the upthrust as shown in the diagram below.


Problem

The upthrust force acting on the wing of aeroplane is given to us. Velocity of the air on its lower surface and area of cross section is also given to us and we need to find the velocity of air on its upper surface ad the problem is as shown in the diagram below.


Solution

We can assume that the wing is horizontal and hence there is no difference in the gravitational energy. As velocity is different kinetic energy per unit mass and hence pressure energy are going to be different. We can apply Bernoullie’s theorem, we can solve the problem as shown in the diagram below.


Problem

Due to a disease main artery expanded in its area of cross section as per the given data in the problem. Velocity of the blood in non expanded area is given to us and blood density is also given to us. We need to find the excess pressure developed due to this and the problem is as shown in the diagram below.
  

Solution

We can apply equation of continuity and velocity of the other part of artery as shown in the diagram below. Now knowing the velocity, we can apply conservation of energy concept and solve the problem as shown in the diagram below.


Problem

The coefficients of viscosity of two liquids is given to us as 2:3 densities of the fluids is given to us as 4:5. In the equal time we need to know the rate of fluid flow in the tubes and the problem is as shown in the diagram below.


Solution

We need to use a concept called Poisellie’s equation. As per it we can write the equation for the rate of flow as shown in the diagram below and solve the problem. Length and diameter of tubes is given as same and the problem is solved as shown in the diagram below.



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Motion in One Dimension Problems with Solutions Twelve

We are interested in solving the problems based on displacement,velocity and acceleration. They are all belongs to one dimensional motion. We know that the velocity is the rate of change of displacement and acceleration is the rage of change of velocity.

In the first problem it is given that position of the particle is given in the XY plane and it is given in terms of time and trigonometric function. We need to know the path of the particle and the problem is as shown below.


Solution

Position vector is given to us in terms of both X and Y components. By multiplying with I unit vector we identify it as the X component. Similarly by multiplying with J, we can identify the component as Y component. J is a unit vector along Y axis. Unit vector is a vector with specific direction and has magnitude of only one unit.

Using trigonometric rule, we can find the relation between X and Y as shown in the diagram below. The path is a circle as the equation supports that mathematically.


Problem

A particle moves according to the equation given in the problem with some constants. We need to find the velocity of the particle in terms of time and we can assume that the body starts from rest. The problem is as shown in the diagram below.


Problem

Rate of change of velocity is given in the problem and we can rearrange the terms to get the component of velocity. But we need the total velocity and to get that we shall integrate the component of velocity with certain given limits. By simplifying the equation further, we can solve the problem as shown in the diagram below.


Problem

A particle moves in a straight line and the position of the particle is given to us in terms of time as shown in the diagram below. We need to know how the velocity and acceleration of the particle changes with respect to time and we need to check with the options given in the diagram below.


Solution

What is given in the problem is in terms of displacement and to get the velocity we need to differentiate the equation. It can be found further that after two seconds velocity of the particle and its displacement becomes zero.

By differentiating the velocity further we can get acceleration as shown below. It can be concluded basing on the equations that if time is less than three seconds, acceleration could be negative.


Problem

It is given in the problem that after the engine is switched off, the boat goes for retardation and it is given as shown in the diagram below. We need to know the velocity of the particle after a specified time.


Solution

By rearranging the terms, we can get a part of velocity. To get the total velocity in the given limits, we need to integrate the equation. Thus we get the velocity of the particle at a given time as shown in the diagram below.



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Motion in One Dimension Problems with Solutions Seven

We are further solving the problems in one dimensional motion. We are here measuring the motion of the body considering only along one direction. In the given problem a body is projected with a known velocity and it reaches a maximum height when thrown vertically up. If the body is thrown with a same velocity and thrown with a known angle so that maximum height is three the range. We need to measure the maximum height of the problem. The problem is as shown in the diagram below.

We can solve one dimensional problem, we can use four equations of motion. They can be used to find the relation among initial velocity, final velocity, acceleration, displacement and time.




Solution

In the given problem maximum height and range relation is given. Maximum height is the vertical height reached by the body and range is the horizontal distance travelled by the body. By substituting the relation we can find the angle of the projection. By further using this value, we can find the maximum height as shown in the diagram below.


Problem

In the given problem two bodies are thrown from the same point with the same velocity and it is given in the problem. It is further given that the angles are complimentary to each other. The difference between the maximum heights of the two projectile is given in the problem. We need to measure the maximum heights of the projectile.


Solution

We know the equation for the maximum height and by putting the values given in the problem, we can find the value as  shown in the diagram below. The angles are zero degree and ninety degree. We can find the angle and then further substituting the value in the maximum height equation, we can find the values as shown in the diagram below.



Problem

It is given in the problem, a player kicks a foot ball with an angle and the velocity of the projection is given in the data. There is another player at a certain distance and he started moving with the kick of the ball. The ball takes time of flight to reach  the ground and the second person has the same time to catch the ball. We need to measure the constant speed with which the player has to run to catch the ball before it hits the ground.



Solution

It is given in the problem that the body is projected with forty five degree angle so that the range becomes maximum.Its value is found to be 40 meter and the second player is at 24 meter. Thus he need to run the distance of 16 meter and he has to run to that distance with a constant velocity and he has the time similar to time of flight of the projectile.


Problem

A body is projected with a known velocity so that it just clears two wall of equal height and the separation between the walls is given to us in the problem. We need to measure the time interval between the two walls to cross.


Solution

Height is the displacement along Y axis. Here we shall take the velocity only along that direction. We can write a two dimensional equation using the second equation of motion and write the equation of the time as shown in the diagram below. The equation has two solutions for the time and they are the times during which the stone cross the two walls. By writing the equation, we can solve the problem as shown in the diagram below.



In the first part of the problem, we had found the difference between them. By simplifying them further, we can solve the problem as shown in the diagram below.



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Motion in One Dimension Problems with Solutions Six

One dimensional motion is the study of the motion of a body only along one direction. In the given problem, a boy sees a ball going up and then down through the window of known height. The total time of visibility the window is given to us in the problem. We need to measure the height above the window. The problem is as shown in the diagram below.

Equations of motion help us in understanding the motion of the one dimensional body. If any three parameters of them are known, by using he appropriate relation, we can find the forth physical quantity.




Solution

As the total time of the journey is one second the upward or downward journey takes half of that time and that is half second. As it is all ready  crossing a certain height h before window, by the time it reaches the window, it will acquire a velocity and we can find that velocity using the third equation of motion.

By substituting that data in the second equation of the motion, we can measure the height above the window as shown in the diagram below.


Problem

In this problem, we are dealing with a body thrown vertically upward with a known velocity. While it is crossing the top of the tower, a man tries to catch it and failed to catch it and it has gone  over crossed above him. Any way in the return journey after three seconds, the ball reached the same person. The boy is able to catch it in the return journey and we are here to measure the height of the tower where that man is standing. The problem is as shown in the diagram below.



Solution

As the velocity of vertical projection is given to us, we can measure the time of ascent. The man is at the maximum height of the tower and he tries to catch it. It is obvious that the tower is at a height less than the maximum height that the stone can travel. He failed to catch it there and it has gone above him further. It is given that it returned to the same place after three seconds. It means it has travelled up further 1.5 second and started returning back. Thus the time for which it has crossed the tower is only one second. By substituting that data in the height of the tower formula, we can find the height of the tower as shown in the diagram below.



Problem

In the given  problem, two stones are thrown vertically upward with the same velocity. But the second stone is thrown after three seconds of the first stone. We need to measure where do these two stones are going to meet ?
Solution

As the bodies are moving against the gravity, we need to treat the acceleration due to gravity as negative. As the times are different by three seconds, we need to take the second stone time as first stone time and less than three seconds as shown in the diagram below. But the height travelled by them has to be same so that they can meet at a certain height. By equating these equations, we can  get the time after which the two stones are going to meet and by further substituting that time data in any one of the height equation, we can measure the height of the meeting point as shown in the diagram below.


Problem

In the given problem, a velocity and time graph is given. We need to measure the ratio of distance travelled by the body in the first two seconds when compared with the total distance travelled by the body in the total time of the journey.



Solution

In the velocity and time graph, we can find the distance in the two seconds can be found as the area between the two seconds. Similarly we can also  find the the total distance as the total area of the graph in the total time as shown in the diagram below.


Problem

In this problem, a ball is thrown vertically up from the ground with a known velocity. At the same time ,another body is thrown down from a known height.It is given in the problem that after some time the two bodies got the collision and here we need to find the velocity of the each body at the time of meeting .


Solution

For the vertically thrown body, acceleration due to gravity shall be treated as negative and for the vertically thrown down body it is positive. For the second body its initial velocity is zero. Equating the equations of both the cases, we can find the specific time after which the stones are going to meet. Again by using that time value, we can find velocities of the bodies using the first equation of the motion as shown in the diagram below.




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