# How to effectively solve One dimensional Motion Problem

This tour is given to give the feel of the whole chapters plus the type of questions and ways to tackle One dimensional motion/linear motion problems.This is quite beneficial for anybody studying 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 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. Magnitude of Instantaneous velocity is equal to instantaneous 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 acceleration as the velocity is changing.

Most Effective way to solve a one dimensional  motion/linear motion  problems

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 of linear motion problems are
v=u+at
s=ut+(1/2)at2
v2=u2+2as
$x=\int v dt$
$v=\int a dt$
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

1. 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
$x=\int v dt$
$v=\int a dt$
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

Some other  good questions for your Assignments

a) 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.

b). 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?

Important Graph in One dimensional Motion

A) Position Time Graph

This is a  plot of Position and time of an object moving along straight line. We can find out velocity,speed,distance and displacement.  We can determine acceleration sign from it but cannot determine the value of it

i) Velocity is given by the Slope of the Position time Graph. Velocity is positive if the slope is positive  and Velocity is negative if the slope is negative

ii) Numerical value of the Slope with out sign gives the speed of the object

iii) Displacement can be calculated using position at the time interval

iv) Displacement time graph is same as Position Time graph  .Distance time can be drawn from position time graph by the taking the mirror of the image of the position-time graph from the point zero velocity onwards

iv) Slope of the chord drawn to displacement time or distance time graph gives the average velocity or average speed  over the time interval to which the chords corresponds to

v) If the graph is concave up, acceleration is + positive and If the graph is concave down, acceleration is -ve

B) Velocity Time graph

i) Area under the velocity time curve provides the displacement

ii) Slope of the curve provides the acceleration

C) Acceleration -time Curve

i) Area under the curve provides the change in velocity

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