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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Best work

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