Equations of Motion
Category : 9th Class
The motion of the body moving along a straight line with uniform acceleration can be described by three equations of motion. These equations can be derived as follows.
First Equation of Motion
Let us consider an object moving with a initial velocity 'u'. If it is subjected to a uniform accleration 'a' such that it attains a velocity of V after time 't', then
Acceleration \[=\frac{Final\,velocity\,-\,initial\,ve\operatorname{lo}city}{Time\,taken}\]
So, \[a=\frac{v-u}{t}\]
\[\Rightarrow \,\,at=v-u,\] or \[v=u+at\]
where, v = final velocity of the body
u = Initial velocity of the body
a = acceleration
and, t= time taken
Second Equation of Motion
The second equation of motion is: \[s=ut+\frac{1}{2}\,a{{t}^{2}}\]. It gives the distance traveled by a body in time t.
Let us derive this second equation of motion:
Let us consider an object moving with a initial velocity 'u' and a uniform acceleration "a". Let it attains a velocity V after some time 't'. Let the distance traveled by the object in this time be 's'. The distance traveled by a moving body in time 't' can be found out by considering its average velocity. Since the initial velocity of the body is 'u' and its final velocity is V, the average velocity is given by:
Average velocity \[=\frac{\text{Initial velocity }+\text{ final velocity}}{2}\]
i.e., Average velocity \[=\frac{u+v}{2}\]
Also, Distance traveled = Average velocity \[\times \] Time
so, \[S=\,\left( \frac{u+v}{2} \right)\times t\] …(1)
From the first equation of motion we have, v = u + at. Putting this value of ‘v’ in equation (1), we get:
Or \[S=\,\left( \frac{u+v}{2} \right)\times t\]
Or \[S=\,\left( \frac{2u+at}{2} \right)\times t\]
Or \[S=\frac{2ut+a{{t}^{2}}}{2}\]
Where s = distance traveled by the body in time t
u = Initial velocity of the body and, a = Acceleration
Third Equation of Motion
The third equation of motion is: v- u^ 2as. It gives the velocity acquired by a body in traveling a distance 's'.
The third equation of motion can be obtained by eliminating -f from the first two equations of motion and using the second equation of motion
From the second equation of motion we have:
\[s=ut+\frac{1}{2}\,a{{t}^{2}}\] …(1)
And from the first equation of motion we have :
\[v=u+at\]
Or, \[at=v-u\]
Or, \[t=\frac{v-u}{a}\]
Putting this value of t in equation (1), we get:
\[S=u\,\left( \frac{v-u}{a} \right)+\frac{1}{2}a\,{{\left( \frac{v-u}{a} \right)}^{2}}\]
Or \[S=u\,\left( \frac{v-u}{a} \right)+\frac{1}{2}\,\frac{{{(v-u)}^{2}}}{a}\]
Or \[S=\frac{uv-{{u}^{2}}+{{v}^{2}}+{{u}^{2}}-2uv}{2a}\]
Or \[2as={{v}^{2}}-{{u}^{2}}\]
Or \[{{v}^{2}}={{u}^{2}}+\,2as\]
Where, v = final velocity,
u = initial velocity,
a = acceleration
and s = distance traveled
Graphical Method of Finding Equations of Motion
We can derive the equation of motions using the velocity time graph. Consider the motion of the object moving along a straight line with a uniform acceleration. The above graph shows the motion of an object along a straight line with a uniform accleration 'a'. The point A and B on the graph corresponds to time 0 and t respectively.
Let u be the velocity of the object at time t = 0 represented by line OA on the graph. Also let the velocity of the particle be vat time t represented by line OD on the graph.
First Equation of Motion
The slope of the velocity-time graph gives the acceleration of an object moving along a straight line.
For line AB, the slope is given by
Slope \[=\frac{BC}{AC}\]
\[\Rightarrow \,\,a=\frac{BC}{AC}=\frac{BE-CF}{OE}\]
\[\Rightarrow \,\,a=\frac{v-u}{t-0}=\frac{v-u}{t}\]
\[\Rightarrow \,\,v=u+at\]
\[\mathbf{=v=u+at}\]
Second equation of Motion
The area under the velocity-time graph is equal to the displacement. In the time interval 0 - t, displacement = area OABE
S = area OABE = area of the rectangle OACE + area of the triangle ABC
\[\Rightarrow \] \[S=OA\times OE+\frac{1}{2}\times AC\times BC\]
\[\Rightarrow \] \[S=OA\times OE+\frac{1}{2}\times AC\times (BC-CE)\]
\[\Rightarrow \] \[S=u\times t+\frac{1}{2}\times t\times (v-u)\] (using first equation of motion)
\[\Rightarrow \] \[S=u\times t+\frac{1}{2}\times a\times {{t}^{2}}\]
or \[\Rightarrow \,\,S=ut+\frac{1}{2}\,a{{t}^{2}}\]
Third equation of motion
The area under the \[v-t\] graph is =area of trapezium AOEB Area of trapezium \[=\frac{1}{2}\] (sum of parallel sides) \[\times \] (altitude)
Or, Area = \[=\frac{1}{2}\times \,(OA+BE)\times AC\]
Or, \[S=\frac{1}{2}\times (u+v)\times t\]
Or, \[S=\frac{1}{2}\times (u+v)\times \,\left( \frac{v-u}{a} \right)\]
Or, \[{{v}^{2}}={{u}^{2}}+2as\]
Average, velocity for uniformly Accelerated Motion
The average velocity of an object can be derived using the first and second equation of the motion. Let us consider an object moving along a straight line with a uniform accleration 'a'. The average velocity can be defined as,
Average velocity \[=\frac{\text{Total Displacement}}{\text{Total Time}}\]
Or, \[\frac{S}{t}=\frac{{{v}^{2}}-{{u}^{2}}}{2at}\]
\[{{V}_{av}}=\frac{v+u}{2},\] (using first equation of motion)
An object moving with a velocity of 15 m/s decelerate at the rate of 1.5 m/ s2 Find the time taken by the objects to come to rest.
(a) 10 sec
(b) 9.5 sec
(c) 11 sec
(d) 12.2 sec
(e) None of these
Answer: (c)
Explanation
We have, from first equation of motion
\[t=\frac{v-u}{a}\]
Or, \[t=\frac{0-15}{-1.5}=\frac{15}{1.5}=10\,\,\sec \]
A car accelerates from 15 km/h to 60 km/h in 300 seconds. Find the distance traveled by the car during this time.
(a) 3.35km
(b) 3.33km
(c) 4.33km
(d) 5km
(e) None of these
Answer: (d)
Explanation
We have, u = 15 km/h, v = 60 km/h and \[t=300\] \[\sec =\frac{1}{12}\]
From first equation of motion,
\[v=u+at,\]
At \[t=0,\] the velocity is u = 20 km/h
At t = 4 min \[=\frac{1}{15}h,\] the velocity is v = 80 km/h
Using \[v=u+at,\]
\[a=\frac{v-u}{t}=a=\frac{60-15}{\frac{1}{12}}\]
Now the distance covered by the car is given by,
\[S=ut+\frac{1}{2}a{{t}^{2}}\]
\[S=15\times \frac{1}{12}+\frac{1}{2}\times 540\times \frac{1}{12}\times \frac{1}{12}=5\,\text{km}\]
Circular Motion
Where, T is the time taken by the object to move around the circular path.
Uniform Circular Motion
When a body moves along a circular path, with constant speed then its direction of motion (or direction of speed) keeps changing continuously.
Circle can be considered to be polygon with infinite sides.
Since the velocity changes due to continuous change in direction, therefore the motion along a circular path is said to be accelerated. When a body moves in a circular path with uniform speed (constant speed), its motion is called uniform circular motion..
An artificial satellite is moving in a circular orbit of radius 32,000 km. If it takes 30 hours to complete one revolution around the earth, then find the velocity of the satellite.
(a) 9428.57 km/h
(b) 6704.76 km/h
(c) 9500 km/h
(d) 7256.68 km/h
(e) None of these
Answer: (b)
Explanation
We have, R =32,000 km and t = 30 hours
Circumference of the orbit \[=2\pi \,\,r\,b=\,2\times \frac{22}{7}\times 32,000=201,142.86\,\text{km}\]
Now, velocity \[=\frac{2\pi R}{T}=\frac{201142.86}{30}=\,6704.76\,km/h\]
A cyclist takes 180 second to complete one round of the circular track. A the radius of the circular track is 90 metres, then calculate his speed. (Given \[\pi =\frac{22}{7}\])
(a) 3.14 m/sec
(b) 3.56 m/sec
(c) 4.25 m/sec
(d) 5.14 m/sec
(e) None of these
Answer: (a)
Explanation
We have \[t=180\,\,\sec \] and \[r=90\] metres
\[V=\frac{2\pi r}{t}\]
Or, \[V=\frac{2\times \frac{22}{7}\times 90}{180}=3.14\,\,m/\sec \]
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