Showing posts with label displacement. Show all posts
Showing posts with label displacement. 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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Equations of Motion along straight line and Freely Falling Video lesson

Equations of motion in one dimensional motion

We know that a body in one dimensional motion has displacement, velocity which is different at initial level and final level and hence the body is also having some acceleration. The body is covering some displacement in a specified time. Taking this into consideration, we can obtain the relation between the above mentioned physical quantities using the equations of motion. They relate some of the above mentioned physical quantities and the relation is among the four quantities. If we know any of the three, by using appropriate equation, we can find the unknown physical quantity using the relations available to us.




 Equation of motion for a freely falling body

A body starting from the state of rest from a certain height and falling vertically downwards under the influence of gravitational force is called as a freely falling body. For a freely falling body, initial velocity is zero and the displacement of the body is nothing but the height of fall in a given time. As the body falls down, its velocity increases and hence it is under acceleration and it is due to gravitational force. This acceleration due to the earth is called acceleration due to gravity and it is constant at a given place. Taking this into consideration, we can rewrite the equations of motion as shown in the video lecture below. The time taken by the body to reach the ground from the maximum height is called time of descent.



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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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Motion in One and two Dimensions Problems with Solutions Fourteen

We are solving problems on one and two dimensional motion in this series of chapters. Here we are dealing with motion of a body moving along only one dimension and also we are studying the motion of a body in a plane where it has motion both along X and Y axis simultaneously.

In the given problem a body is dropped from a certain height and it strikes the inclined plane at a height h above the ground. As a result of impact, velocity of the body becomes horizontal. We need to measure  that it will take maximum time to reach the ground basing on what condition. The problem is  as shown in the diagram below.


Solution

It is given that the body is dropped from a certain height and hence its initial velocity is zero. It has fallen only some distance before it actually strikes the ground. That can be found as the difference between the total height from where it  is dropped and the location where it hit the ground.

Thus we can measure the time of fall in both the cases using the equations of motion. We need to measure the total maximum journey time. For the parameter to be maximum, its differentiation function with height has to become zero as a mathematical rule. 

By simplifying it, we can find that relation as shown in the diagram below.


Problem

It is given in the problem that a projectile is projected with a velocity and a known angle of projection. It is given that the trajectory grazes the vertices of the triangle as shown in the diagram. We need to know the relation between the angles with the horizontal.



Solution

The diagram for the problem data is as shown in the figure. We can express the angles with the horizontal in terms of horizontal and vertical displacement as shown in the diagram below.

We also know that the vertical displacement can be expressed in terms of horizontal displacement and angle of projection in terms of range as shown further.

By comparing this two equations, we can find the relation as shown below.


Problem

Two cannons are installed at the top of a cliff of 10 meter and they fire a shot with a known speed at some interval of time. The first cannon is fired with an angle and the other one is done horizontally. We need to know where do the two cannons collide with each other. The problem is as shown in the diagram below.


Solution

Both of them were fired from a certain height. But one horizontally and the other with an angle. By substituting the given data, we can simplify the equations as shown in the diagram below.



Problem

It is given in the problem that a particle is projected from the ground with a initial speed at angle of projection with the horizontal. We need to measure the average velocity of the projection between its point of projection and the highest point of the trajectory. The problem is as shown in the diagram below.


Solution

We know that the average velocity is the ratio of total displacement of a body in the total time of the journey. We can find the total displacement of the journey as a vector sum of horizontal and vertial displacements and the time of journy is half of time of flight.By substituting that data, we can  solve the problem as shown in the diagram below.



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Motion in One Dimension Problems with Solutions Two

We would like to solve the first problem basing on one dimensional body. The problem is about a body thrown initially with zero velocity. After two seconds another body is dropped with a certain velocity. We need to know the velocity of that body so that both the bodies reach the ground simultaneously.

Problem One


Solution

The first body is initially has zero velocity and using the second equation of motion, we can find the time taken by the body to reach the bottom of the bridge and strike the water. It is given that the second stone is thrown down with a initial velocity after two seconds. So the second body takes only three seconds to reach the ground. Both the stones reach the bottom of the bridge so that the distance travelled by both of them is same.

By substituting them in the same second equation of motion, we can find the initial velocity with which the second stone is thrown from the same place as shown in the diagram below.


Problem Two

This problem of one dimensional motion along Y axis is also of the similar  model of the previous problem. But in this problem, both the stones are falling freely so that have no initial velocity. The second stone is thrown few seconds later and we would like to know the distance of separation between the stones has to be a certain value after a specified time and we would like to find it out.


Solution

As per second equation of motion, we have relation between displacement, initial velocity, acceleration and time as shown in the diagram below. The second equation is different from the first as there is a time gap between both the stones.

It is given that after a certain time, we need to have a certain difference of separation. Thus by subtracting them we can solve the problem as shown in the diagram below.


Problem Three

This problem is also about two stones falling from a certain height and the second stone is having a falling distance certain separation 10 meter when compared with the first one.

It is given in the problem that both of them reach the bottom of the tower at the same time and the second stone is actually thrown one second later than the first one.


Solution

For a freely falling body, by using the second equation of motion, we can find the height travelled by the body  before it had reached the bottom. For the second stone also, we shall apply the same equation, but the time is one second less and height is 10 meter less.

By substituting the value of the height from the first equation in the second equation, we can solve the problem as shown in the diagram below.We can find both the time as well as height of the tower.


Problem Four

The problem is about the distance travelled by the body in the particular second. We have forth equation of motion in one dimensional motion. It is given that in the given second a certain distance of seven meter and in the previous second five meter has fallen. We would like to measure the velocity of the body with which it strikes the ground.


Solution

By solving them using forth equation of motion as shown in the diagram below after dividing both the cases, we can get the value of the n th second. Further we can find the final velocity of the body using the first equation of motion. The solution is shown in the diagram below.


Problem Five

It is given in the problem that one stone is dropped from a certain height 180 meter. It has no initial velocity. Another stone is dropped from a certain point below the first height and it means, it is travelling less distance than the first one. Both of them reached the ground at the same time but the second one got started initially late by two seconds.


Solution

By using the second equation of the motion, we can write two equations as shown in the diagram below. By solving both the equations, we can solve the problem as shown in the diagram below.


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Horizontal Projectile, Applications Problems with Solutions

A body projected horizontally from a certain height with an initial horizontal velocity can be called as a horizontal projectile. Its initial velocity along the vertical direction is zero and it possess only horizontal velocity at the beginning. As the time progresses, due to the impact of the gravity, it acquires the vertical component of velocity also. It can be shown that path taken by this body is parabola using the equations of motion of kinematics. We can write the equations for the displacement along x-axis and y-axis and further substituting the value of the time from the x-axis equation in the y-axis equation, it can be proved that its path is parabola.

At any instant of the path, velocity of the projectile is tangential to its path. It has both horizontal and vertical components and the angle made by the velocity vector with the horizontal can be calculated as shown below. We can also calculate the effective velocity of the projectile at that instant.


Application

Let us consider a body projected horizontally from the top of a tower. The line joining the point of projection and the striking point of ground makes an angle of 45° with the horizontal.  What is the displacement of the body?

Solution

As the angle made at the horizontally is 45 degree, according to trigonometry, its horizontal and vertical distances are equal. Being the displacement is a vector quantity with can calculate its resultant value using the parallelogram law of the vectors as shown below.


Application

Let us consider a ball is thrown horizontally from a staircase as shown with a initial velocity. We shall calculate how many steps it travels before it strikes the ground?

Solution

The ball is having only horizontal component of velocity and has no initial vertical component of velocity.

We can use equations of motion to find the displacement along x-axis and y-axis. There is no acceleration due to gravity impact on the x-axis and there is gravity acting along the y-axis. The total horizontal distance travelled by the body is equal to the multiplication of the number of the steps and the breadth of each step. Similarly the total vertical distance travelled by the body is equal to the multiplication of number of steps with the height of each step.

By substituting this values in the above equation is can solve the problem as shown below.


Application:

Two bodies are thrown horizontally with a two different initial velocities in mutually opposite directions from the same height. What is the time after which velocity is of the two bodies are perpendicular to each other ?

Solution:

Velocity is a physical quantity that is having both magnitude and direction and it has to be treated like a vector quantity. When two vectors are perpendicular to each other their scalar product becomes zero. Taking this into consideration and by writing the equations of velocity is for the two bodies after a specific time and equating their product to 0 with can solve the problem as shown below.


Problem

An object is thrown horizontally from a point it hits the ground at some another point. The line of sight from these two points makes an angle 60° with the horizontal. If acceleration due to gravity is 10 m/s Squire and time of flight is known what is the velocity of projection?

Solution

by writing the equations were displacement along x-axis and y-axis and by using a trigonometrical definition we can solve the problem as shown below.


Problem

Two paper screens are separated by a distance of hundred meter. A bullet fired through them. The hole made in the second screen is 10 cm below the hole made in the first screen. What is its initial velocity before it strikes the first screen?

Solution

Initially the bullet is having only horizontal velocity and its vertical component of velocity equal to 0. There is no impact of acceleration due to gravity along the x-axis and it is acting along the y-axis. Taking this things into consideration we can equate the distance between the two poles along the vertical direction is equal to the vertical distance travelled by the body during the journey time. We can also calculate the time taken by the body for this to happen using the equation of the displacement along the x-axis. By solving these equations together we can calculate the initial velocity of the projection as shown below.


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Problems on Motion of a Body Along a Straight Line

A body moving along only one direction during its motion is said to be in one-dimensional motion. It is also the motion along a straight line when the air friction is neglected. To apply the laws of motion, we can consider a particle concept. Particle is a body of negligible dimensions. It is of point size and having negligible mass. In mechanics we generally derive all the laws and formulas for this particle. As the body is the combination of identical particles, the concepts that are generated for the particle can be applied to the body also at the broader level.

To study the motion of the body along one-dimension, we have four equations of motion. If the body is moving along the horizontal direction, it may have acceleration which is nothing but rate of change of velocity. If the body is moving along the vertical direction, it is moving against or in favor of the gravity. In that case acceleration of the body is called acceleration due to gravity and it is a constant value which is equal to 9.8 m/s Squire.

The relation among initial velocity, final velocity, acceleration, time of travel and the distance is discussed in the previous post and we are going to use the same relations to solve these problems.

Problem 

A bus accelerates from rest at constant a rate for some time, after which it starts de accelerates at another constant rate to come to the state of rest. If the total time allotted for the journey is given, calculate the maximum speed and the total distance traveled by the bus?

Solution:

Let us divide the total time of the travel into two parts and when we add these two parts where going to get the total time. To solve this problem we can draw a graph taking the time on x-axis and the velocity on y-axis. The graph starts from the origin and is a straight line until it reaches its maximum velocity during the positive acceleration period. After reaching the peak, as per the problem the particle starts negative acceleration and hence the graph also starts decreasing in the reverse way and finally reaches the x-axis as shown.

We know that the slope of the velocity and the time graph gives the acceleration. The first part of the slope is going to be a positive slope because it is an increasing slope and the second part of the graph is a negative slope because it is decreasing. Anyway we can consider the magnitude of the slope to solve the problem.

By identifying the slope and further by identifying the time taken for the journey, we can calculate the total time as shown below.

We also know that the area of velocity and time graph gives the displacement. By identifying the area of the diagram, we can also calculate the total displacement as shown in the attached diagram.



Alternate solution

We can solve the same problem without using the graphical concept. Here also let us assume that a body is accelerated for a specific time, reaches its maximum velocity, and then starts de accelerating and finally comes the state of rest after a specified time. The sum of these two specified times gives the total time of the journey and it is given in the problem.

By using the first equations of motion we can calculate the time of travel and by adding that two equations, we can get the total time as shown below. From the velocity maximum equation, we can calculate the acceleration as the effective value of both acceleration and de acceleration and by substituting the value in the second equation of motion, we can also get the total displacement as shown below.



Problem 

A body starts from rest and travel is a distance with a uniform acceleration. Then it moves further some distance with a uniform velocity and after covering the specified distance, starts de accelerating and comes to the state of rest after covering some different distance. We need to find out the ratio of average velocity to the maximum velocity?

Solution

We can again draw a graph taking the time and x-axis and the velocity on y-axis. Initially the graph starts from the origin and it is a straight line and reaches a point where the body acquires a maximum velocity. As this velocity is going to remain constant for the second part of the displacement, the graph is going to be a straight line parallel to the x-axis. 

After covering that specified second part of the distance, the body starts de accelerating, the graph also starts reversing itself to the x- axis as shown. Every time taking the concept that the area of velocity and the time graph is equal to displacement, we can get the equations for individual times of the three different parts of the problem.

We know that the average velocity is defined as the ratio of total displacement to the total time and by substituting the respective values we can calculate the ratio of average velocity to the maximum velocity as shown below.



Problem 

In the arrangement shown in the diagram, the ends of in extensible string moves in the downward direction with the uniform speed. The pullies in the given situation are fixed. Find the speed with which the mass moves in the upward direction?

Solution

By looking at the diagram, it can be understood that the length is a constant value and its differentiation with respect to time is going to give zero. This is the basic rule of the differentiation that differentiation of any constant is equal to 0. We can assume that the body is moving up with a certain displacement and correspondingly with a certain velocity. By identifying the relation between positions and by differentiating the positions equation once with respect to time we can get the velocities as shown below.



Problem 

The diagram shows a Rod of a specified length resting on a wall in the floor. Its lower end is pulled two words the left direction with a constant velocity. Find the velocity of the other end of rod moving in the downward direction, where the rod is making a specific and a known angle with the horizontal?

 Solution 

The situation the problem is as shown in the diagram. We can again read the relation between lengths of the Rod, the distance of one end of the rod from the origin along the x-axis and the distance of the second end of the Rod from the origin along the y-axis.

As we are in need of velocity, differentiating the displacement equation is with respect to time, we can get them as shown below.




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