{"id":1472,"date":"2020-06-10T03:58:00","date_gmt":"2020-06-09T22:28:00","guid":{"rendered":"http:\/\/physicscatalyst.com\/article\/?p=1472"},"modified":"2022-11-04T12:06:28","modified_gmt":"2022-11-04T06:36:28","slug":"solve-motion-in-one-dimension-problems","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/article\/solve-motion-in-one-dimension-problems\/","title":{"rendered":"How to effectively solve Motion in One dimension Problems"},"content":{"rendered":"\n

Motion in one dimension is one of the important chapters in Mechanics. This post is given to give the feel of the whole chapters plus the type of questions and how to solve motion in one dimension problem. This is quite beneficial for anybody studying One-dimensional Motion.<\/p>\n\n\n\n

\"how<\/a><\/figure>\n\n\n\n

Description:<\/strong><\/span>
One dimensional Motion is the study of the motion along a straight line. The complete Study material has been provided at the below link
motion in one dimension<\/a><\/p>\n\n\n\n

Most Important Points to remember about One dimension Motion<\/strong><\/span>
1. Distance and displacement are not the same things. 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 its normal position, its displacement is zero while the distance traveled is the circumference of the circle
2. Speed is calculated over distance while velocity is over displacement. So Average speed in a round trip will never be zero while average velocity will be zero
3. The magnitude of Instantaneous velocity is equal to the 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 the change in velocity per unit time. A body moving with constant speed but with varying directions will have acceleration as the velocity is changing.<\/p>\n\n\n\n

Most Effective way to solve a motion in straight line problems<\/strong><\/span><\/p>\n\n\n\n

1.<\/strong> First, visualize the question
2. Take starting point as the origin and take one direction as positive and others 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. if the motion is varying and the relation is given for that motion. Then we can utilize a one-dimensional motion derivative equation to find out the solutions
5. Calculate as require<\/p>\n\n\n\n

Formula’s of linear motion problems are<\/strong><\/span>
$v=u+at$
$s=ut+  \\frac {1}{2}at^2$
$v^2=u^2+2as$
$x=\\int v dt $
$v=\\int a dt $
$v= \\frac {dx}{dt}$
$a=\\frac {dv}{dt}= \\frac {d^2x}{dt^2}$
$a=v \\frac {dv}{dx}$<\/p>\n\n\n\n

Example -1<\/strong><\/span>
A bus starts at Station A from rest with a uniform acceleration of 2m\/sec2<\/sup>. The bus moves along a straight line
1. Find the distance moved by bus in 10 sec?
2. At what time, its velocity becomes 20 m\/sec?
3 How much time it will take to cover a distance of 1.6 km<\/p>\n\n\n\n

Solution:<\/strong><\/span>
Now first step to attempt such a question is to visualize the whole process. Here the bus is moving along a straight line and with uniform acceleration
Now what we have
Initial velocity=0
Acceleration=2m\/sec2<\/sup><\/p>\n\n\n\n

Now since it is uniform motion we can use given motion formula’s in use
$v=u+at$
$s=ut+  \\frac {1}{2}at^2$
$v^2=u^2+2as$<\/p>\n\n\n\n

(1) distance (s)=?
time(t)=10 sec
So here the most suitable equation is
s=ut+(1\/2)at2<\/sup>
Substituting given values
s=(0)10 +(1\/2)2(10)2<\/sup>=100 m<\/p>\n\n\n\n

(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<\/p>\n\n\n\n

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

Example -2<\/strong><\/span>
An object is moving along a straight line. The motion of that object is described by
$x=at+bt^2+ct^3$
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<\/p>\n\n\n\n

Solution:<\/strong><\/span>
Now first step to attempt such a 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
$x=at+bt^2+ct^3$
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 $
$v= \\frac {dx}{dt}$
$a=\\frac {dv}{dt}= \\frac {d^2x}{dt^2}$
$a=v \\frac {dv}{dx}$<\/p>\n\n\n\n

(1) x=? t=1sec<\/p>\n\n\n\n

Here by substituting t=1 in given equation we get the answer
x=a(1)+b(1)2<\/sup>+c(1)3<\/sup>
x=a+b+c m<\/p>\n\n\n\n

(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<\/sup>+ct3<\/sup>)\/dt
or v=a+2bt+3ct2<\/sup>
Substituting t=0 we get
v=a m\/s
Substituting t=1 we get
v=a+2b+3c m\/s<\/p>\n\n\n\n

(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<\/sup>)\/dt
a=2b+6ct
Substituting t=0 we get
a=2b m\/s2<\/sup>
Substituting t=1 we get
a=2b+6c m\/s2<\/sup><\/p>\n\n\n\n

Some other  good questions for your Assignments<\/strong><\/span><\/p>\n\n\n\n

(a) The displacement of the body x(in meters) varies with time t (in sec) as
x=(-2\/3)t2<\/sup> +16t+2
find following
a. what is the velocity 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 be when it comes to rest
e . How much time it takes to come to rest.<\/p>\n\n\n\n

(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 a constant acceleration of 1.2 m\/s2<\/sup>
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 be running at the same speed ever catch up the bus?<\/p>\n\n\n\n

Some Videos on how to solve motion in one dimension problem<\/strong><\/span><\/p>\n\n\n