How to solve Motion Problems of 9th Class

How to solve Motion Problems of 9th Class

The Motion problem in Class 9 are limited to Straight line motion 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 Motion for class 9.

Complete Study material has been provided at the below link

[standout-css3-button href=”https://physicscatalyst.com/Class9/motion.php”]Class 9 Motion notes[/standout-css3-button]

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 possesses uniform velocity if it covers equal distance in same time in specified 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 amount 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 motion problem

1. Read the problem statement carefully 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. Usually the object moving direction is chosen 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 initial velocity is zero. Both the initial and final values of quantities 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 motion ,then you can utilize the One dimensional motion 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

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

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

3. 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 known and unknowns Initial velocity=0 ( as car starts from rest) Acceleration=1m/sec²

Fourth step :Now since it is uniform acceleration motion we can use given motion 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

1. 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 motion which is a special case for Rectilinear motion

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 downwards

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 known and unknowns Initial 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 motion we can use given motion formula’s in use

v=u+at

s=ut+(1/2)at²

v²=u²+2as

Fifth steps : Carry out the calculation

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

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 appropriate

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 quantities defined in this Article

[standout-css3-button href=”https://physicscatalyst.com/Class9/motion_cm.php”]Motions Concept Map[/standout-css3-button]

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

[standout-css3-button href=”https://physicscatalyst.com/Class9/motion_questions.php”]Motions Assignments[/standout-css3-button]

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