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### One Dimensional Motion Tour

This tour is given to give the feel of the whole chapters plus the type of questions and ways to tackle tehm.This is quite beneficial for anybody studing One dimensional Motion.

Description:
One dimensional Motion is the study of the motion along an straight line.Complete Study material has been provided at the below link
Study Material

Most Important Points to remember about One dimension Motion
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 cirle and return to it normal position,it displacement is zero while the distance tarvelled is the circumference of the circle
2. Speed is calculated over distance while velocity over displacment.So Average speed in a round trip will never be zero while average velocity will be zero
3. Magnitude of Instantanous velocity is equal to instantanous speed
4. When a particle moves with constant velocity, its average velocity, its instantaneous velocity and its speed are all equal
5. Acceleration is defined as change in velocity per unit time.A body moving with constant speed but with varying direction will have accleration as the velocity is changing.

Most Effective way to solve a one dimensional motion problem

1. First 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
3. Write down what is given in the question and what is required
4. If it is uniform motion ,then you can utilize the One dimensional motion equation.It the motion is varying and relation is given for that motion.Then we can utilize one dimensional motion derivative equation to find out the solutions
5. Calculate as require

Formula's are
v=u+at
s=ut+(1/2)at2
v2=u2+2as
x=∫vdt
v=(dx/dt)
a=dv/dt=(d2x/dt2)
a=vdv/dx

Example -1
A bus start at Station A from rest with uniform acceleration 2m/sec2.Bus moves along a straight line
1.Find the distance moved by the bus in 10 sec?
2.At what time,it velocity becomes 20m/sec?
3 How much time it will take to cover a distance of 1.6km

Solution:
Now first step to attempt such question is to visualize the whole process.Here the bus is moving along a straight line and with uniform acceleration
Now what we have
Intial velocity=0
Acceleration=2m/sec2

Now since it is uniform motion we can use given motion formula's in use
v=u+at
s=ut+(1/2)at2
v2=u2+2as

1. distance (s)=?
time(t)=10 sec

So here the most suitable equation is
s=ut+(1/2)at2
Substituting given values
s=(0)10 +(1/2)2(10)2=100 m

2. velocity(v)=20 m/s
time(t)=?
So here the most suitable equation is
v=u+at
20=0+2t
or t=10 sec

3.distance(s)=1.6km=1600m
t=?
So here the most suitable equation is
s=ut+(1/2)at2
1600=(1/2)(2)t2
or t=40 sec

Example -2
A object is moving along an straight line.The motion of that object is described by
x=at+bt2+ct3
where a,b,c are constants and x is in meters and t is in sec.
1. Find the displacement at t=1 sec
2. Find the velocity at t=0 and t=1 sec
3. Find the acceleration at t=0 and t=1 sec

Solution:
Now first step to attempt such question is to visualize the whole process.Here the object is moving along a straight line and its motion is described by the given equation
Now we
x=at+bt2+ct3
Now since its motion is described by the given equation,following formula will be useful in determining the values
x=∫vdt
v=(dx/dt)
a=dv/dt=(d2x/dt2)
a=vdv/dx

1. x=? t=1sec

Here by substituting t=1 in given equation we get the answer
x=a(1)+b(1)2+c(1)3
x=a+b+c m

2.v=? t=0,v=? t=1
Here we are having the displacement equation,so first we need to find out the velocity equation
So here the most suitable formula is
v=(dx/dt)
or v=d(at+bt2+ct3)/dt
or v=a+2bt+3ct2
Substituing t=0 we get
v=a m/s
Substituing t=1 we get
v=a+2b+3c m/s

3 a=? t=0,a=? t=1
Now we are having the velocity equation,we need to first find the acceleration equation.
So here the most suitable formula is
a=(dv/dt)
or a=d(a+2bt+3ct2)/dt
a=2b+6ct
Substituing t=0 we get
a=2b m/s2
Substituing t=1 we get
a=2b+6c m/s2

First check your concepts for this chapter by attempting the questions given at the link.Solutions are also provided there.
Conceptual Test

Once you are comfortable with the concept,you are ready to go for full length questions of various types.Please follow the below link one by one .Give your full try and check out the solutions.

Objective Test
Subjective Test
Graphical Questions Test

Once you are comfortable with these questions,I will suggest to revise all the concepts once again so that these concepts becomes embeded in your mind.