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Equation of motion is a mathematical formula that describes how a physical system behaves over time. The equation of motion describes objects and systems' motion in terms of dynamic variables. These equations relate various important parameters of motion such as velocity, displacement, speed, time, and acceleration.
- An object is said to be in motion if it changes its position with respect to the same reference point or frame of reference with time.
- Motion can be described in two ways: Dynamics and Kinematics
- Dynamics deals with the study of the motion of the bodies taking into account the forces which cause the motion in the bodies.
- Kinematics deals with studying the motion of the bodies without taking into account the cause of the motion in the bodies.
- At constant acceleration, the kinematic equations of motion are referred to as the SUVAT equations, arising from the definitions of kinematic quantities: displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t).
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Key Terms: Motion, velocity-time graph, acceleration, displacement, velocity, time, speed, movement, rotational motion, force, distance.
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Motion
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Motion is a phenomenon in which the body changes its position with respect to the same reference point or frame of reference with time.
- If the position of an object changes with respect to one particular point then the object is considered to be in a state of motion with respect to that point.
- To understand the state of motion, there are some equations that are derived by using the entities like displacement, movement, velocity, speed, and acceleration.
- The relation between these terms is often called the equations of motion.
Equations of Motion
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There are a total of three equations of motion that can be derived by algebraic, graphic, and calculus methods. Equations of motion apply to uniformly accelerated motion. These equations are as follows:
- First Equation of Motion: v = u + at
- Second Equation of Motion: s = ut + 1/2 at2
- Third Equation of Motion: v2 = u2 + 2as
Where
- v is the final Velocity
- u is the initial velocity
- t is the time taken
- a is constant acceleration or uniform acceleration
- s is the displacement of the body
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First Equation of Motion
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The First equation of motion is known as the velocity-time relation and it is given by
v = u + at
Where,
- v = final velocity of the body
- u = initial velocity of the body
- t = time taken to change the velocity of the body
- a = constant acceleration
Derivation of First Equation of Motion by Algebraic Method
Consider a particle moving in a straight line with constant acceleration(a). Let v and u be the velocities of the body time t = t and t = 0 respectively.
The acceleration of the body is given by
\(a={change\, in\, velocity\over time\, taken}={v-u\over t-0}\)
\(\Rightarrow a={v-u\over t}\)
Rearranging the above equation, we get
v = u + at
Derivation of First Equation of Motion by Graphical Method
From the graph, we have
- OA = CD = u
- BC = (v – u)
- OD = t
The slope of the v-t graph gives acceleration, therefore
\(a=slope\,of\,v-t\,graph=tan\theta={BC\over AC}={BC\over OD}\)
\(\Rightarrow a={v-u\over t} \Rightarrow v=u+at\)
Derivation of First Equation of Motion by Calculus Method
By the definition of acceleration, \(a={dv\over dt} \Rightarrow dv=a\,dt\)
Integrating both sides, we get
\(\int dv=\int a\,dt\)
As acceleration is constant, therefore it comes out of the integral.
Let at t = 0, the initial velocity of the particle is u, and its final velocity at time t is v
\(\Rightarrow \int_u^v dv=a\,\int_0^tdt \Rightarrow v-u=a(t-0)\)
\(\Rightarrow v=u+at\)
Second Equation of Motion
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The Second equation of motion is known as the position-time relation and is given by
\(s=ut+{1\over2}at^2\)
Where,
- s = displacement of the body
- u = initial velocity of the body
- t = time taken
- a = constant acceleration
Derivation of Second Equation of Motion by Algebraic Method
Consider a particle moving in a straight line with constant acceleration(a). Let x and xo be the position of the body at time t = t and t = 0 respectively. Let v and u be the velocities of the body at time t = t and t = 0 respectively.
The average velocity of the body is given by
\(v_{av}= {(x-x_o) \over(t-o)} \Rightarrow x-x_o =v_{av}t\)
But, \(v_{av}={v+u\over 2} \Rightarrow x-x_o =({v+u\over 2}) t\)
But, v = u + at, Hence
\(x-x_o =({u+at+u \over 2})t\)
\(\Rightarrow x-x_o =ut+{1\over 2}at^2\)
Let x – xo = s, displacement of the body
\(\Rightarrow s =ut+{1\over 2}at^2\)
Derivation of Second Equation of Motion by Graphical Method
From the graph, we have
- xo = position of the particle at t = 0.
- x = position of the particle at t = t.
- x – xo = s = displacement of the particle in time t
But s = area under v-t graph = area of trapezium OABD
⇒ s = 1/2[sum of parallel sides].[distance between parallel sides]
⇒ s = 1/2[OA + BD].[AC] = 1/2[u + v]t
But, v = u + at
⇒ s = 1/2[u + u + at]t = 1/2[2u + at]t
\(\Rightarrow s=ut+{1\over 2}at^2\)
Derivation of Second Equation of Motion by Calculus Method
By the definition of velocity, \(v={dx\over dt} \Rightarrow dx=v\,dt\)
In the above equation, v is not independent of t. So we replace the value of v by v = u + at.
\(\Rightarrow dx=(u+at)dt\)
Integrating both sides, we get
\(\int dx=\int (u+at)dt\)
Let at t = 0, x = xi (initial position)
And, at t = t, x = x (final position)
\(\Rightarrow \int_{x_i}^x dx=\int_0^t (u+at)dt\)
On solving the above equation, we get
\(x-x_i =ut+{1\over 2}at^2\)
But x – xo = s = displacement of the particle in time t
\(\Rightarrow s=ut+{1\over 2}at^2\)
Third Equation of the motion
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The Third equation of motion is known as the velocity-position relation or velocity-displacement relation and is given by
\(v^2=u^2+2as\)
Where,
- v = final velocity of the body
- u = initial velocity of the body
- a = constant acceleration
- s = displacement of the body in time t
Derivation of Third Equation of Motion by Algebraic Method
From the first equation of motion, we have
\(v=u+at \Rightarrow t={v-u \over a}\) ...(i)
Also from the second equation of motion, we have
\(s=ut+{1\over 2}at^2\)
Substituting equation (i) in the above equation, we get
\(s=u({v-u \over a})+ {1\over 2}a({v-u \over a})^2\)
On solving the above equation, we get
\(v^2=u^2+2as\)
Derivation of the Third Equation of Motion by Graphical Method
From the graph, we have
s = area under v-t graph = area of trapezium OABD
⇒ s = 1/2[sum of parallel sides].[distance between parallel sides]
⇒ s = 1/2[OA + BD].[AC] = 1/2[u + v]t ...(i)
From the first equation of motion, we have
\(v=u+at \Rightarrow t={v-u \over a}\) ...(ii)
Substituting equation (ii) in equation (i), we get
\(s={1\over2 }(v+u)({v-u\over a})\)
\(\Rightarrow s={v^2-u^2\over 2a}\)
\(\Rightarrow v^2=u^2+2as\)
Derivation of Third Equation of Motion by Calculus Method
By the definition of acceleration, \(a={dv\over dt}\)
We can also write it as
\(a={dv \over dx}.{dx \over dt}\)
But, dx/dt = v, therefore
\(a=v{dv\over dx} \Rightarrow v\,dv = a\,dx\)
Integrating both sides with limits, we get
\(\int_u^v v\, dv= \int_{x_i}^x a\,dx\)
On solving the above equation, we get
\(v^2-u^2 = 2a(x-x_i)\)
But x – xo = s = displacement of the particle in time t
\(\Rightarrow v^2-u^2 = 2as\)
\(\Rightarrow v^2 = u^2+2as\)
the particle in time t
\(\Rightarrow s=ut+{1\over 2}at^2\)
Equation of Rotational Motion
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To describe the rotational motion of the body, we will need certain parameters:
- The angle of rotation(θ): It is defined as a measure of the amount of rotation of the body.
- Angular velocity(ω): It is defined as the rate of rotation. The angular velocity is defined as:
\(\omega = {d\theta\over dt}\)
- Angular acceleration(α): It is defined as the rate of change of angular velocity and it is given by
\(\alpha = {d\omega\over dt}\)
Equations of Rotational Motion for constant angular acceleration are given by:
- ω = ω0 + αt
- θ = ω0t + 1/2 αt2
- ω2 = ω02 + 2αθ
Applications of the Equations of Motion
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- It is used to calculate the kinetics problems.
- By identifying other parameters, we use the equations to calculate the parameter which isn't given.
- Used to calculate optical properties.
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Things to Remember Based on Equations of Motion
- Equations of Motion can be derived with the help of the velocity-time graph.
- Motion is defined as the change in position or place of an object.
- If the position of an object changes with respect to a point then the object is considered to be in the state of motion with respect to that point.
- Acceleration is defined as a rate of change in the velocity of a moving object.
- The first equation of motion is - v-u = at.
- The second equation of motion is- s= ut + ½ at2
- The third equation of motion is v2 = u2+ 2as
Important Questions Based on Equations of Motion
Ques: The method that is used for the derivation of the equation of motion is known as: (1 Mark)
(A) Algebraic method
(B) Graphical method
(C) Calculus method
(D) All of these
Ans: B) Graphical method
Ques: An object dropped from a cliff falls with a constant acceleration of 10 m/s2. Find its speed 5 s after it was dropped. (1 Mark)
Ans: By the first equation of motion-
v=u+at
Here u=0, a=10 m/s2, t= 5 s
v=10×5=50m/s
Ques: A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4 m. Determine the acceleration of the bike. (2 Marks)
Ans: Given that-
u = 0 m/s
v = 7.10 m/s
s = 35.4 m
a = ?
According to the Third Equation of Motion:
v2 = u2 + 2as
(7.10)2 = (0)2 + 2*(a)*(35.4)
50.4 = (70.8)*a
(50.4)/(70.8) = a
a = 0.712 m/s2
Ques: An engineer is designing the runway for an airport. Of the planes that will use the airport, the lowest acceleration rate will likely be 3 m/s2. The takeoff speed for this plane will be 65 m/s. Assuming this minimum acceleration, what is the minimum allowed length for the runway? (3 Marks)
Ans: Given that-
u = 0 m/s
v = 65 m/s
a = 3 m/s2
s = ?
As per the Third Equation of Motion
v2 =u2+ 2as
(65)2 = (0)2 + 2*(3)*s
4225 = (6)*s
(4225)/(6) = s
d = 704 m
Ques: A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.840 m. Determine the acceleration of the bullet (assume a uniform acceleration). (2 Marks)
Ans: According to the Third Equation of Motion:
v2 = u2+ 2as
(521)2 = (0)2 + 2*(a)*(0.840)
271441 = (1.68 m)*a
271441/1.68 = a
a = 1.62 x 105 m /s2
Ques: The observation deck of a tall skyscraper 370 m above the street. Determine the time required for a penny to free fall from the deck to the street below. (2 Marks)
Ans: As per the second equation of motion
s= ut + ½ at2
-370 = (0)*(t)+ 0.5*(-9.8)*(t)2
-370 = 0+ (-4.9)*(t)2
(-370)/(-4.9) = t2
75.5 = t2
t = 8.69 s
Ques: An object moves along a straight line with an acceleration of 2m/s2. If its initial speed is 10 m/s, what will be its speed 2 s later? (2 Marks)
Ans: Here u=10m/s, a=2m/s2, t = 2 sec
By the first equation of motion-
v =u+at
v=10+2×2 =10+4 =14m/s
Ques: A train starts from rest and accelerates uniformly at the rate of 5 m/s2 for 5 sec. Calculate the velocity of the train in 5 sec. (2 Marks)
Ans: Given u=0, a= 5m/s2, t= 5 sec, v=?
By the first equation of motion-
v=u+at
v=0+5×5
v=25 m/s
Ques: A train accelerates from 26 km/h to 42 km/h in 10 sec.
(i) Acceleration
(ii) The distance travelled by train. (3 Marks)
Ans: (i) Acceleration is given by
a=Δv/Δt
Δv=42−26=16km/hr=16×1000/3600 m/s=4.4m/s
So a=4.4/10=.44 m/s2
(ii) By the second equation of motion-
s=ut+1/2at2
Now u=26 km/hr =7.22m/s
s=7.22×10+1/2×.44×102=72.2+22
Distance traveled, s=94.2 m
Ques: A scooter traveling at 15 m/s speeds up to 25 m/s in 4 sec. Find the acceleration of the scooter. (3 Marks)
Ans: Initial velocity (u) = 15m/s
Final velocity (v) = 25 m/s
Time (t) = 4s
By First Equation of Motion
v = u + at
So, a = (v-u)/t
or, a = (25-15)/4 =10/4 =2.5 m/s2
Ques: An electron moving with a velocity of 5× 103m/s enters into a uniform electric field and acquires a uniform acceleration of 103 m/s2 in the direction of the initial velocity.
i) Find out the time at which electron velocity will be doubled?
ii) How much distance would the electron cover in this time? (3 Marks)
Ans: Given u=5× 103m/s, a=103 m/s2, v=2u=10×103m/s
i) By Using
v=u+at
10×103=5× 103+103t
or
t=5 sec
ii) By Using the Second Equation of Motion
s=ut+1/2at2
s=5× 103×5+1/2×103×52
=37.5×103m
Ques: A car starts from rest and acquires a velocity of 54 km/h in 2 sec. Find
a) the acceleration
b) distance travelled by car (assume motion of the car is uniform) (3 Marks)
Ans: Here,
Initial velocity = u = 0 m/s
Final velocity = v = 54 km/hr = 15 m/s
Time = t = 2 sec
By first equation of motion-
v = u + at
Here a = (v-u)/t = 15/2 = 7.5 m/s2
By the second equation of motion:
s = ut + 1/2 at2 = 0 + 1/2 × 7.5 × 22 = 15 m
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