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.

1 comment:

  1. can you put problems of ray optics having both mirror and lens in the same problem????

    ReplyDelete