How to solve Motion Problems of 9th Class



How to solve Problems of 9th Class

The problem in are limited to Straight line with either no acceleration or constant acceleration.
This article is given to give the feel of the whole chapters plus the type of questions and ways to tackle them.

This is quite beneficial for anybody studying for .

Complete Study material has been provided at the below link

Class 9 Motion notes

First we need to remember these important terms

Distance Distance in the length of pathn describe by the object moving in space.
It is scalar quantity. It SI unit is meter
Displacement Displacement is the shortest distance between initial and final object position with direction.
It is vector quantity. It SI unit is meter
Average Speed Average speed is defined as the total distance covered divided by total time to cover the total distance.
It is scalar quantity. It SI unit is meter/sec. A body is suppose to have constant speed if it cover
same distance in same time
Average Velocity Average velocity is defined as the total displacement covered divided by total time to cover the total displacement. It is vector quantity. It SI unit is meter/s.A body posseses uniform velocity if it covers equal distance in same time in specfied direction
Acceleration Acceleration is defined as rate of change of velocity per unit time.
It is vector quantity. It SI unit is meter/sec 2.
A body will have uniform acceleration if it travels in straight line and velocity increases by same ammount in equal interval of time

Some important concepts to remember and understand

1. Distance and displacement are not the same thing.Distance is scalar while displacement is a vector quantity.Distance is never zero in a round trip while the displacement will be zero.Example When a person moves in a circle and return to it normal position,it displacement is zero while the distance travel led is the circumference of the circle

2. Speed is calculated over distance while velocity over displacement.So Average speed in a round trip will never be zero while average velocity will be zero

3. When a particle moves with constant velocity, its average velocity and average speed is equal equal

4. Acceleration is defined as change in velocity per unit time.A body moving with constant speed but with varying direction will have acceleration as the velocity is changing.

 

Most Effective way to solve the problem

1. Read the problem statement carelfully and visualize the question

2. Take starting point as origin as and take one direction as positive and other as negative.This is required as we will be dealing with Vector quantities. Ususally the object moving direction is choosen as the positive direction

3. Write down what is given in the question and what is required. You may have to translate the stated languages in physics terms like started at rest means intial velocity is zero. Both the intial and final values of quantites are written if given

4) Now we need to find what principles of physics is required to solve the problem and what all formula’s to be used

if it is zero acceleration problem, then we can simply the equation

s=ut where s is displacement and U is velocity and t is time

If it is uniform acceleration ,then you can utilize the One dimensional equation.

v=u+at

There are four variable in this equation,So if we know any three variable, we can get the fourth one. If acceleration,time and initial velocity is know, we can find the final velocity from this

s=ut+(1/2)at2

There are four variable in this equation,So if we know any three variable, we can get the fourth one. If acceleration,time and initial velocity is know, we can find the total displacement

v2=u2+2as

There are four variable in this equation,So if we know any three variable, we can get the fourth one. If acceleration,displacement and initial velocity is know, we can find the final velocity

5. Carry out all the calculation and find out the answer.

6. Think carefully about the answer ,if it looks reasonable

7. Checks for the units

Example -1

A car start from rest with uniform acceleration 1m/sec2 along a straight line

1.Find the distance moved by the car in 5 sec?

2.At what time,it velocity becomes 20m/sec?

3 How much time it will take to cover a distance of .8 km

Solution:

First step :Now first step to attempt such question is to visualize the whole process.Here the car is moving along a straight line and with uniform acceleration

Second Step : Here the car is moving in one direction,so let us assume that as positive direction

Third step : We need find out the knowns and unknowns Intial velocity=0 ( as car starts from rest) Acceleration=1m/sec2

Fourth step :Now since it is uniform acceleration we can use given formula’s in use

v=u+at

s=ut+(1/2)at2

v2=u2+2as

Fifth steps : Carry out the calcultaion

1. distance (s)=? time(t)=5 sec

So here the most suitable equation is s=ut+(1/2)at2



Substituting given values s=(0)5 +(1/2)1(5)2=12.5 m

2. velocity(v)=20 m/s time(t)=?

So here the most suitable equation is

v=u+at

20=0+1t

or t=20 sec

3.distance(s)=0.8km=800m t=?

So here the most suitable equation is

s=ut+(1/2)at2 800=(1/2)(1)t2

or t=40 sec

Sixth step: The answer looks reasonable.Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. It is very clear from data from above that RHS is equal to LHS

Seventh step: We have taken correct units

Now lets us take one example for Free fall which is a special case for Rectilinear

First of all some facts about Free fall acceleration

1) A acceleration g=9.8 m/s2 acts downward on the object when either it is thrown upward or fallen downward from the top building

2) The acceleration g is same for all the objects irrespective of the mass of the body in absence of air friction.

3) It always act downward

Example 2:

Roxxan throws a ball of mass=.5 kg upward in the air with a velocity of 20m/s. Assume negligible air friction g=10 m/s2

1) Find out the time to reach there

2) Find out the height to which the ball will reach

Solution:

First step :Now first step to attempt such question is to visualize the whole process.Here the ball is moving along a straight line in upward direction and with uniform acceleration direction downwwards

Second Step : Here the ball is moving in one direction,so let us assume that upward direction as positive direction

Third step : We need find out the knowns and unknowns Intial velocity=20 m/s ,final velocity=zero( ball will be at rest at the top point) Acceleration=-10m/sec2( negative as it is directed downward)

Fourth step :Now since it is uniform acceleration we can use given formula’s in use

v=u+at

s=ut+(1/2)at2

v2=u2+2as

Fifth steps : Carry out the calcultaion

1. v=0,u=20 m/s,a=-10 m/s2 time(t)=? The below equation would be appropiate

v=u+at

0=20-10t

t=2 sec

2) v=0,u=20 m/s,a=-10 m/s2 s=?

The below equation would be appropiate

v2=u2+2as

0=400-20s s=20 m

We need to Practice many questions to gain confidence and expertise in this concept.

Do check these concept Map for a good understand of important physical quanties defined in this Article

Motions Concept Map

There are lot of assignment present at this link. Do attempt them

Motions Assignments

If you are interested in other links containing notes in science and maths along with free download links consider following the links given below
Class 9 science study material Class 9 maths study material Class 9 downloads



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