## Monday, February 23, 2015

### 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.

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