Showing posts with label distance. Show all posts
Showing posts with label distance. 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.


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.


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.


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.


Acceleration due to gravity and One Dimensional Motion Equations

Acceleration due to gravity

The Earth is a huge massive body. According to Newton law of gravitation, the force of attraction between the two bodies is directly proportional to the product of the masses and inversely proportional to the square of the distance of separation. Here the two masses involved are the mass of the body and the mass of the Earth. The distance between the bodies is generally radius of the Earth, as the body is on the surface of the earth or near to the surface of the earth. As the mass of the earth is comparatively very high when compared with the body, the gravitational force of attraction will be always towards the Centre of the Earth. This gravitational force provides acceleration to all the bodies on the Earth and the corresponding acceleration is called acceleration due to gravity. The numerical value of acceleration due to gravity on the Earth is 9.8 m/s Squire. It may vary slightly from place to place but in general it can be taken as a constant.

This acceleration on the surface of the earth is uniform and constant. Hence when the body is coming towards the Earth, because of acceleration due to gravity its velocity will be keep on increasing until it strikes the ground. If you throw the body against the acceleration due to gravity, into the sky, it’s velocity will be keep on decreasing and finally it stops at a certain height called maximum height. Time taken to reach that maximum height is called time of ascent and time taken to reach the ground from a certain height is called time of descent.

We have four different equations of the motion to express the translatory motion of a body. In all that equations where ever acceleration is there, we can substitute acceleration due to gravity and we can rewrite the corresponding equations of motion due to gravity. Here there are practically two possibilities. The body may be coming towards the Earth or the body may be going away from the Earth. The acceleration due to gravity on the bodies that are coming towards the earth is generally treated as positive and vice versa.

We can write the equations of motion and the corresponding values as shown below.


Using this equations we can find the time taken by the body to reach a particular height against the gravity and it is called time of ascent. If the body is falling from the same height to reach the ground to take the same time and that is called time of dissent. The sum of the time of ascent and time of dissent is called time of flight.

A body is said to be a freely falling body if it is falling from a certain height with zero initial velocity. In that case, using the equations of motion we can calculate the final velocity of the body after covering a certain height h.

It can be also proved that if your body is thrown up with the velocity from the ground, after reaching the maximum height it will come back to the ground with the same velocity but in the reverse direction.


In general we have ignored the impact of the air resistance while measuring the time of ascent and time of descent. If we consider the air resistance, then the time of ascent is going to be little bit different from time of dissent. The direction of the force due to the air resistance is always in opposition to the relative motion.

So the effective acceleration in this case is going to be more than acceleration due to  gravity and hence the time of ascent is less than that of the case when the air resistance is ignored. When we are ignoring air resistance we are imagining that the environment is vacuum for the calculation purpose. Though it is not practically vacuum the given equations will be approximately valid in almost all the real-time situations.

By taking their resistance into consideration if we try to calculate the time of dissent, being their resistances against the relative motion, it is in the upward direction but the gravity is in the downward direction. And hence the effective acceleration will be less and hence time of dissent will be more and more than the case of vacuum.

It can be also be noticed that time of decent is more than the time of recent when their resistances taken into consideration.


Problem and solution

Two balls are dropped from different heights. One ball is drop two seconds after the other but both the strikes the ground at the same time which is five seconds after the first ball. Find the heights from where these two balls dropped?

As the first stone is taking five seconds to reach the ground and the second stone is two seconds late, it shall be only taking three seconds to reach the ground. The corresponding equations for the heights of the bodies can be written basing on the equation of motion as shown below. Simply the difference between the heights of the two equations we can calculate it as shown below.


Problem and solution

A body balls freely from a height of 125 m. After two seconds gravity ceases to act. Find the time taken by it to reach the ground if acceleration due to gravity is considered as 10 metre per second Squire.

The body is a freely falling body means its initial velocity equal to 0. For the first two seconds acceleration due to gravity is acting and hence it falls due to the gravity and it covers a distance of 20 m. During this process the body will acquire some velocity and it can be calibrated using the equations of motion as shown below. It is found that its value equal to 20 meter per second. Being it is drop from 125 m and 20 m is covered in the first two seconds, it has to further cover a distance of 105 m per reach the ground.

As acceleration due to gravity is no more acting the velocity of the body will remain constant. Does the body will continue to move with the same velocity of 20 m/s for the remaining time. We can calculate the time taken to cover the 105  m is in the formula that the displacement equal to velocity multiplied by time. By adding this time of the two seconds with can get the total time of the journey.



Problem and solution

A parachutist drops freely from an aeroplane for 10 seconds before the parachute opens out. He’s velocity when he reach the ground is 8 m/s. Once if the parachute is open he has a standard retardation of two meter per second Squire. Find the height at which he gets out of the aeroplane.

Before the parachute opens he falls down due to the gravitational effect and he’s not having a initial velocity in the vertically downward direction. Hence we can find that is going to cover a distance of 490 m in the first 10 seconds. In this 10 seconds will also acquire some velocity due to acceleration due to gravity and it can be found that the velocity is 98 m/s.

He covers the further distance with the standard retardation of two meter per second Squire and we can calculate the distance that he covers with a initial velocity of 98 m/s and a final velocity of 8 m/s and with an acceleration of two meter per second Squire using the equation of motion. By substituting the value is we can get the total height as 2875 m.




Related Posts

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.

Equations of Motion in One Dimension

To explain the motion of a body along one-dimension, we have four equations of motion. The parameters like initial velocity, final velocity, time, acceleration and displacement varies in one-dimensional motion and in these relations, and we are going to find the appropriate relation between these physical quantities under different circumstances. Any of these equations are simple relation between any four of the above-mentioned physical quantities.

If we know any of the three physical quantities, we can find out the Forth physical quantity by using appropriate relation.

Each of the equation and its meaning is as mentioned below.



Graphical method to prove equations of motion

If a graph is drawn taking the displacement on x-axis and time on y-axis, the slope of the graph gives the velocity. If a graph is drawn taking the velocity on x-axis and the time on y-axis, the slope of the graph gives acceleration. The area of this graph gives the total displacement. We can calculate the area of the graph depending on the shape of the graph.

Here we are interested to calculate the equations of motion using this kind of graphical method. Here time is taken on x-axis and the velocity is taken on y-axis. The body is already having some initial velocity and hence the graph is not going to start from the origin. As the body is having a uniform velocity the graph is rather a straight line making a constant angle with horizontal from a given point. After some time it acquires a final velocity V and it is represented in the graph as shown. By writing the definition of the slope of the graph, we can derive the first equations of motion as shown below.


We can derive the second equation of motion with the help of the same above graph. This can be done by taking the area of the graph covered under the given velocity time graph. As the area covers is a rectangular and triangle, by writing the area of these two parts with can get the total displacement as shown below.


Problem and solution

The speed of a train is reduced from 60 km/h to 15 km/h while travelling a distance of 450 m. If its retardation is uniform, find how much further distance it  travel before it comes to the state of rest?

While solving this problem we can use the equations of motion. Initial velocity is and final velocity is given but they are given in terms of kilometres per hour. As we are solving the problem in the standard international system, we shall convert them into metre per second. This can be converted by multiplying with the term 5/18.

By using the third equation of motion in the first part of the problem, we can find the acceleration of the body and this acceleration is uniform. That means even in the second part of the motion it will continue to have the same retardation. By taking that acceleration into consideration and by applying the equation of motion again to the second part of the problem, we can calculate the further distance travelled by the body as shown below.


Problem and solution

A car is moving with the velocity of 20 m/s. The driver observes stationary truck ahead at a distance of hundred meter. After some reaction time, the brakes are applied by the driver and he produced the retardation of 4 m/s Squire. What is the maximum reaction time that he can have so that he can avoid the collision?

It takes some time for any of the human being, to understand that he has to apply the brake after seeing an obstacle. This time is called reaction time. In the meantime he will be continuing to move with the same uniform velocity and hence he’ll be covering some distance.

Then he applies the break with a constant retardation and he further has to cover some more distance so that he has to stop before he reach the obstacle. It is noticed that, basing on the third equation of motion, he will cover a distance of 50 m once if you apply the brake. So he can cover a distance of 50 meter before he apply the brake therefore he can avoid the collision. Hence we can travel with the same constant velocity of 20 m/s for a distance of 50 m and he take a time of 2.5 seconds for the. The detailed solution is as mentioned below.


Problem and solution

The bus starts moving with an acceleration of two meter per second Squire. A Boy is 96 m behind the bus and start simultaneously running with the velocity of 20 m/s. After what time, he will be able to catch the bus?

As both of them are started simultaneously, the boy has to cover two distances. One is the gap between him and the bus and the other is the distance travelled by the bus itself in a given time.

To calculate the distance travelled by the bus we can use the second equation of motion. As the boy is moving with a constant velocity, the distance covered by him is nothing but the product of velocity and time.

A detailed solution is as mentioned below. You can catch the bus twice once after eight seconds and once more after 12 second. If that 12 second is crossed he’ll be never able to catch the bus again.





Related Posts