Kinematics problem

Question
A car accelerates from rest at a constant rate p for some time after which it decelerates at a constant rate q to come to rest. If total time lapse is t, find (1) the maximum velocity attained and (2) the total distance traveled.

Solution

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Inertial and non inertial frames

• Both rest and motion are relative terms and there is nothing like absolute rest or absolute motion.
• Position or state of motion of a body may appear different in different frame of reference i.e.,  an object at rest in one frame of reference might appear to be in motion in another frame of reference.
• Newton's first law of motion also known as law if inertia does not always holds for all frame of reference and the frames in which this law holds good are called inertial frames of reference.
• Inertial frames of reference are non-accelerating frames this means that either they are fixed or move with constant velocity.
•  Non-inertial frames are the frames in which newton's law of inertia does not holds true. 
• Rotating and acceleration frame of references falls in the category of non inertial frames.
• In this frame acceleration is caused by fictitious or pseudo forces.

For more or full notes on kinematics visit this link

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=∫adt
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=∫adt
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.

Graphical Question for Kinematics

1. A ball is dropped vertically from a height h above the ground .It hits the ground and bounces up vertically to a height h/2.Neglecting subsequent motion and air resistance ,its velocity v varies with the height h as






















2. The displacement -time graph of a moving particle is shown below.The instantanous velocity of the particle is negative at the point









a. C
b. D
c. B
d. A

3.The velocity -time graph of a moving particle is shown below.Total displacement of the particle during the time interval when there is nonzero acceleration and retardation is










a. 60m
b. 40m
c. 50m
d. 30m

4. Figure below shows the displacement -time graph of two particles.Mark the correct statement about their relative velocity










a. It first increases and then decreases
b. It is a non zero constant
c. it is zero
d. none of the above

5.Four position -time graph are shown below.What all graph shows motion with positive velocities






















a. a and c only
b all the four
c. b and D only
d. b only

6. What all graph shows motion with negative velocities
a. b and d only
b all the four
c. a and c only
d. c only

7.which of the folliwing graph correctly represents velocity-time relationships for a particle released from rest to fall under gravity





















8.The v-x graph(fig1) of a particle moving along a straight line is shown below.Which of the following below graph(A,B,C,D) shows a-x graph
























Solutions

Kinematics Numerical

1. A boy is standing at a distance a₁ from the foot of a tower.The boy throws an stone at a angle 45° which just touches the top of the tower and strikes the ground at a distance a₂ from the point the boy is standing.Find the height of the tower

2. A ball is thrown upward from a point on the side of a hill which slopes upward uniformly at angle 45°.Intial velocty of the ball is u₀ m/s and at an angle 65°(with respect to the horizotal).At what distance up the slope the ball strike and in what time?


3.A cannon on a level plain is aimed at an angle θ above the horizontal.A shell is fired with a muzzle velocity v₀ towards a pole which is distance R away.It hits the pole at height H.
a find the timetaken to reach the pole
b. find the value of H in terms of θ,R and v₀


4. The displacement of the body x(in meters) varies with time t (in sec) as
x=(-2/3)t² +16t+2
find following
a. what is the velocty at t=0,t=1
b. what is the acceleration at t=0
c. what is the displacement at t=0
d .what will the displacement when it comes to rest
e .How much time it take to come to rest.

5. A man runs at a speed at 4 m/s to overtake a standing bus.When he is 6 m behind the door at t=0,the bus moves forward and continues with constant acceleration of 1.2 m/s²
find the following
a. how long does it take for the man to gain the door
b if in the beginning he is 10m behind the door ,will he running at the same speed ever catch up bus?


Detailed Solutions:
3. The time taken to reach the pole is given
R=v₀cosθ t
or t=R/v₀cosθ

Now equation of motion of vertical motion
h=v₀sinθt-1/2gt²
at t=R/v₀cosθ h=H
so
H=(v₀sinθ)X(R/v₀cosθ ) -(1/2)g( R/v₀cosθ)²
or H=Rtanθ-gR²/2v₀²cos²θ


4.

Given x=(-2/3)t² +16t+2

Velocity=dx/dt=(-4/3)t+16
so velocity at t=0 =16
and velocity at t=1 =(-4/3)+16=44/3

Acceleration is given as =d²x/dt²=-4/3

so acceleration is time independent and it is constant

Displacement at t=0 can be found by simply substituting the values of t=0 in equation (1)
=2

Now v=(-4/3)t+16
when v=0
then t=4/3 sec

Displacement can be found by substituing the value of t=4/3 in equation (1)
=-32/27+64/3+2
=(-32+576+54)/27=598/27 m