The equations of motion are a set of mathematical equations which describe the motion of bodies moving with uniform acceleration. These equations show the relationship between initial velocity, final velocity, acceleration, time taken, and distance travelled by a body during motion. In kinematics, the three equations of motion are extremely important because they help in analysing different types of motion and solving numerical problems in a simple and systematic manner. These equations are widely used in physics, engineering, transportation, and many real-life applications involving moving objects.
Derive the Formulas of Equations of Uniformly Accelerated Motion ?
There are three equations for the motion of those bodies which travel with a uniform acceleration. These equations give relationship between initial velocity, final velocity, time taken, acceleration and distance travelled by the bodies. We will study these equations one by one.
Derive First Equation of Motion : $v=u+at$
The first equation of motion is :
$$v=u+at$$
It gives the velocity acquired by a body in time $t$. We will now derive this first equation of motion.
Consider a body having initial velocity $u$. Suppose it is subjected to a uniform acceleration $a$ so that after time $t$ its final velocity becomes $v$.
From the definition of acceleration, we know that :
$$
\text{Acceleration} = \frac{\text{Change in velocity}}{\text{Time taken}}
$$
or
$$
\text{Acceleration} = \frac{\text{Final velocity} – \text{Initial velocity}}{\text{Time taken}}
$$
Therefore,
$$
a = \frac{v-u}{t}
$$
Multiplying both sides by $t$,
$$
at = v-u
$$
Rearranging,
$$
v = u + at
$$
where:
$v = \text{final velocity of the body}$
$u = \text{initial velocity of the body}$
$a = \text{acceleration}$
$t = \text{time taken}$
The equation $v = u + at$ is known as the first equation of motion. It is used to find out the velocity $v$ acquired by a body in time $t$, when the body has an initial velocity $u$ and a uniform acceleration $a$.
In fact, this equation contains four quantities. If any three quantities are known, the fourth quantity can be calculated.
By paying due attention to the sign of acceleration, this equation can also be applied to the problems of retardation.
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Derive Second Equation of Motion : $s=ut+\frac{1}{2}at^2$
The second equation of motion is :
$$s=ut+\frac{1}{2}at^2$$
It gives the distance travelled by a body in time $t$.
Let us derive this second equation of motion.
Suppose a body has an initial velocity $u$ and a uniform acceleration $a$ for time $t$ so that its final velocity becomes $v$. Let the distance travelled by the body in this time be $s$.
The distance travelled 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:
$$
\text{Average velocity}=\frac{\text{Initial velocity}+\text{Final velocity}}{2}
$$
That is,
$$
\text{Average velocity}=\frac{u+v}{2}
$$
Also,
$$
\text{Distance travelled}=\text{Average velocity} \times \text{Time}
$$
So,
$$
s=\left(\frac{u+v}{2}\right)t \qquad …(1)
$$
From the first equation of motion, we have:
$$
v=u+at
$$
Putting this value of $v$ in equation (1), we get:
$$
s=\left(\frac{u+u+at}{2}\right)t
$$
$$
s=\left(\frac{2u+at}{2}\right)t
$$
$$
s=\frac{2ut+at^2}{2}
$$
$$
s=ut+\frac{1}{2}at^2
$$
where
$s = \text{distance travelled by the body in time } t$
$u = \text{initial velocity of the body}$
$a = \text{acceleration}$
This is the second equation of motion and it is used to calculate the distance travelled by a body in time $t$.
This equation should also be memorized because it will be used to solve numerical problems.
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Third Equation of Motion : $v^2=u^2+2as$
The third equation of motion is :
$$v^2=u^2+2as$$
It gives the velocity acquired by a body in travelling a distance $s$. We will now derive this third equation of motion.
The third equation of motion can be obtained by eliminating $t$ between the first two equations of motion.
From the second equation of motion, we have:
$$
s = ut + \frac{1}{2}at^2 \qquad …(1)
$$
And from the first equation of motion, we have:
$$
v = u + at \qquad …(2)
$$
This can be rearranged and written as:
$$
at = v-u
$$
or
$$
t = \frac{v-u}{a}
$$
Putting this value of $t$ in equation (1), we get:
$$
s = u\left(\frac{v-u}{a}\right) + \frac{1}{2}a\left(\frac{v-u}{a}\right)^2
$$
$$
s = \frac{u(v-u)}{a} + \frac{(v-u)^2}{2a}
$$
Taking L.C.M.,
$$
s = \frac{2u(v-u) + (v-u)^2}{2a}
$$
Using the identity:
$$
(v-u)^2 = v^2 + u^2 – 2uv
$$
we get:
$$
s = \frac{2uv – 2u^2 + v^2 + u^2 – 2uv}{2a}
$$
$$
s = \frac{v^2 – u^2}{2a}
$$
Therefore,
$$
2as = v^2 – u^2
$$
or
$$
v^2 = u^2 + 2as
$$
where
$v = \text{final velocity}$
$u = \text{initial velocity}$
$a = \text{acceleration}$
$s = \text{distance travelled}$
This equation gives us the velocity acquired by a body in travelling a distance $s$.
We will now solve some problems based on motion. To solve the problems on motion we should remember that :
1. If a body starts from rest, its initial velocity is :
$$
u = 0
$$
2. If a body comes to rest (it stops), its final velocity is :
$$
v = 0
$$
3. If a body moves with uniform velocity, its acceleration is :
$$
a = 0
$$
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Derivation for Displacement Travel in nth Second of Its Motion
In kinematics, we often calculate the total displacement of a moving object over a given time. However, it is also useful to find the displacement in a particular second, such as the 5th or 10th second. The formula for displacement in the nth second helps us determine this.
The displacement of a particle moving with uniform acceleration in $n$ seconds is given by :
$$S_n = u n + \frac{1}{2} a n^2$$
where :
$u$ = Initial velocity of particle
$a$ = Acceleration produced in the particle
$n$ = Time in seconds
The displacement in $(n – 1)$ seconds is :
$$S_{n-1} = u (n-1) + \frac{1}{2} a (n-1)^2$$
So, the displacement in the $n$th second of its motion is :
$$D_n = S_n – S_{n-1}$$
Substituting values :
$$D_n = \left( u n + \frac{1}{2} a n^2 \right) – \left( u (n – 1) + \frac{1}{2} a (n – 1)^2 \right)$$
Expanding, $$D_n = u n + \frac{1}{2} a n^2 – u (n – 1) – \frac{1}{2} a (n^2 – 2n + 1)$$
Simplifying, $$D_n = u + \frac{a}{2} (2n – 1)$$
This is the equation to find the displacement in the $n$th second of is motion.
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Solved Numerical Problems Based on Equations of Motion
Q.1 : A scooter acquires a velocity of $36$ km per hour in $10$ seconds just after the start. Calculate the acceleration of the scooter.
Solution :
First of all, we should convert the given velocity into proper units, that is, we should convert the velocity of 36 kilometres per hour into metres per second (because the time is given in seconds).
Now, 1 km = 1000
So, 36 km = 36 x 1000 m = 36,000 m …(1)
Also, 1 hour = 60 minutes = 60 x 60 seconds = 3600 s …(2)
So, 36 km per hour = 36,000 m/3600 s = 10 m/s …(3)
Thus, the given velocity of 36 km per hour is equal to 10 metres per second.
Now, Initial velocity, $u$ = 0 (Scooter starts from rest)
Final velocity $v$ = 10 m/s (Calculated above)
Acceleration, a = ? (To be calculated)
And, t = 10 s
By putting these values in the first equation of motion :
$v = u + at$
10 = 0 + a x 10
10 a = 10
a = 10/10
a = 1 m/s2
Thus, the acceleration of the scooter is 1 m/s2.
Q.2 : A moving train is brought to rest within 20 seconds by applying brakes. Find the initial velocity, if the retardation due to brakes is 2 m s–2.
Solution.
In this problem we have been given the value of retardation but we require the value of acceleration. We know that retardation is negative acceleration. So, if the retardation is, + 2 m s–2 (as given here), then the acceleration will be, – 2 m s–2 (the minus sign here indicates the negative acceleration). Let us solve the problem now.
Here, Initial velocity, $u$ = ? (To be calculated)
Final velocity, $v$ = 0 (The train stops)
Acceleration, $a$ = – 2 m s–2
And, Time, t = 20 s
Now, putting these values in the formula :
$v = u + at$
0 = $u$ + (– 2) × 20
0 = $u$ – 40
$u$ = 40 m s–1
Thus, the initial velocity of the train is 40 metres per second.
Q.3 : A body starts to slide over a horizontal surface with an initial velocity of 0.5 m/s. Due to friction, its velocity decreases at the rate of 0.05 m/s2 (acceleration, – 0.05 m/s2). How much time will it take for the body to stop ?
Solution.
Here, Initial velocity, $u$ = 0.5 m/s
Final velocity, $v$ = 0 (The body stops)
Acceleration, $a$ = – 0.05 m/s2
And, Time, $t$ = ? (To be calculated)
Now, from the first equation of motion, we have :
$v = u + at$
0 = 0.5 + (– 0.05) × $t$
0.05 $t$ = 0.5
$t$ = 10 s
Thus, the body will take 10 seconds to stop.
Q.4 : A racing car has a uniform acceleration of 4 m/s2. What distance will it cover in 10 seconds after the start ?
Solution.
Here, Initial velocity, $u$ = 0
Time, $t$ = 10 s
Acceleration, $a$ = 4 m/s2
And, Distance, $s$ = ? (To be calculated)
Now, putting these values in the second equation of motion :
$s = ut+\frac{1}{2}at^2$
$s = 0\times{t}+\frac{1}{2}at^2$
$s$ = 0 × 10 + $\frac{1}{2}$ × 4 × (10)2
$s$ = 200 m
Thus, the distance covered by the car in 10 seconds is 200 metres.
Q.5 : A scooter moving at a speed of $10\text{ m/s}$ is stopped by applying brakes which produce a uniform acceleration of $-0.5\text{ m/s}^2$. How much distance will be covered by the scooter before it stops?
Solution:
Initial speed ($u$) = $10\text{ m/s}$
Final speed ($v$) = $0\text{ m/s}$ (since the scooter stops)
Acceleration ($a$) = $-0.5\text{ m/s}^2$
Distance covered ($s$) = ? (to be calculated)
According to equation of motion :
$$v^2 = u^2 + 2as$$
Substituting the given values into the equation :
$$(0)^2 = (10)^2 + 2 \times (-0.5) \times s$$
$$0 = 100 – 1.0 \times s$$
$$0 = 100 – s$$
Rearranging the equation to solve for $s$ :
$$s = 100\text{ m}$$
Thus, the distance covered by the scooter before it comes to a stop is 100 meters.
Q.6 : A car travelling at $20\text{ km/h}$ speeds up to $60\text{ km/h}$ in $6\text{ seconds}$. What is its acceleration?
Solution :
Before substituting the values into the formula, we must convert the speeds from kilometers per hour ($\text{km/h}$) to meters per second ($\text{m/s}$) so that all units match the standard SI system.
To convert $\text{km/h}$ to $\text{m/s}$, you multiply by $\frac{1000\text{ m}}{3600\text{ s}}$ or simply multiply by the fraction $\frac{5}{18}$.
Initial speed ($u$) = $20\text{ km/h} = 20 \times \frac{5}{18} = 5.55\text{ m/s}$
Final speed ($v$) = $60\text{ km/h} = 60 \times \frac{5}{18} = 16.66\text{ m/s}$
Time taken ($t$) = $6\text{ s}$
Acceleration ($a$) = ? (to be calculated)
According to equation of motion :
$$v = u + at$$
Rearranging the formula to isolate acceleration ($a$):
$$a = \frac{v – u}{t}$$
Substituting the converted values into the formula :
$$a = \frac{16.66 – 5.55}{6}\text{ m/s}^2$$
$$a = \frac{11.11}{6}\text{ m/s}^2$$
$$a \approx 1.85\text{ m/s}^2$$
Thus, the acceleration of the car is $1.85\text{ m/s}^2$.
Q.7 : A bus increases its speed from $20\text{ km/h}$ to $50\text{ km/h}$ in $10\text{ seconds}$. What is the acceleration of the bus.
Solution :
Before calculating the acceleration, we need to convert the initial and final speeds from kilometers per hour ($\text{km/h}$) to meters per second ($\text{m/s}$) so that all units are in the standard SI system.
To convert $\text{km/h}$ to $\text{m/s}$, multiply the value by $\frac{5}{18}$ (or $\frac{1000\text{ m}}{3600\text{ s}}$).
Initial speed ($u$) = $20\text{ km/h} = \frac{20 \times 1000\text{ m}}{3600\text{ s}} \approx 5.55\text{ m/s}$
Final speed ($v$) = $50\text{ km/h} = \frac{50 \times 1000\text{ m}}{3600\text{ s}} \approx 13.88\text{ m/s}$
Time taken ($t$) = $10\text{ s}$
Acceleration ($a$) = ? (to be calculated)
According to the equation of motion :
$$v = u + at$$
Rearranging the formula to solve for acceleration ($a$):
$$a = \frac{v – u}{t}$$
Substituting the values into the formula :
$$a = \frac{13.88 – 5.55}{10}\text{ m/s}^2$$
$$a = \frac{8.33}{10}\text{ m/s}^2$$
$$a \approx 0.833\text{ m/s}^2$$
Thus, the acceleration of the bus is $0.83\text{ m/s}^2$.
Frequently Asked Questions FAQs on Equations of Motion
What are equations of motion?
Equations of motion are mathematical relations that describe the motion of bodies moving with uniform acceleration. These equations connect important physical quantities such as velocity, acceleration, distance, and time. They are widely used in physics to solve numerical problems related to motion.
Why are equations of motion important in physics?
Equations of motion help us understand how an object changes its speed and position with time. They are useful in solving practical problems involving vehicles, trains, falling objects, and other moving bodies. These equations form the foundation of kinematics in mechanics.
What is meant by uniform acceleration?
Uniform acceleration means that the velocity of a body changes by equal amounts in equal intervals of time. In such motion, the acceleration remains constant throughout the journey. Most equations of motion are applicable only when acceleration is uniform.
What is the first equation of motion used for?
The first equation of motion is used to calculate the final velocity of a moving body after a certain time. It relates the initial velocity, acceleration, and time taken. This equation is especially useful in problems involving speeding up or slowing down.
What does the second equation of motion explain?
The second equation of motion gives the distance travelled by a body during a given time interval. It is commonly used in numerical problems involving displacement and acceleration. This equation is very important for calculating the position of moving objects.
What is the significance of the third equation of motion?
The third equation of motion relates velocity, acceleration, and distance travelled without involving time. It is useful in situations where the time taken is not known. This equation is frequently applied in braking and stopping distance problems.
What is retardation in motion?
Retardation is negative acceleration, which means the speed of a body decreases with time. It acts opposite to the direction of motion and slows the body down. Examples include applying brakes to a car or a train coming to rest.
Why do we convert km/h into m/s in numerical problems?
In physics, SI units are used for consistency and accuracy in calculations. Since time is usually given in seconds, velocity must be converted into metres per second. This ensures that all quantities are expressed in compatible units before applying formulas.
What happens when a body starts from rest?
When a body starts from rest, its initial velocity is considered zero. In such cases, the equations of motion become simpler and easier to solve. Many practical motion problems involve bodies starting from rest.
Can equations of motion be used in daily life?
Yes, equations of motion are used in many real-life situations such as calculating the speed of vehicles, stopping distances, motion of elevators, sports movements, and transportation systems. Engineers, scientists, and drivers all use these concepts directly or indirectly.
Here is a purely text-based FAQ designed to answer common conceptual questions about uniformly accelerated motion without using any mathematical equations.
What does it mean when we say an object is moving with uniform acceleration?
Uniform acceleration means that the speed or velocity of an object changes by exactly the same amount during every equal interval of time. For example, if a car increases its speed by five meters per second during the first second, it will continue to increase its speed by exactly five meters per second during every single second that follows. This indicates that the rate of speeding up or slowing down remains completely constant throughout the entire journey.
How do we handle situations where an object is slowing down or coming to a stop?
When an object slows down, it experiences a specific type of uniform acceleration called retardation or deceleration. In conceptual terms, this just means the speed is decreasing at a steady, fixed rate over time instead of increasing. If the object comes to a complete halt or stops moving entirely, its final velocity simply drops down to zero, which helps define the endpoint of that specific motion.
What variables connect the different equations of motion together?
The equations of motion are designed to link five core aspects of a journey: the starting speed, the ending speed, the total time spent moving, the steady rate of acceleration, and the total distance covered. Each individual equation is built to leave out one of these factors so you can solve a problem depending on what details you already know. If you are tracking an object that starts completely from a standstill or rest, you simply consider its initial starting speed to be zero.
What is meant by displacement in the nth second?
Displacement in the nth second refers to the distance covered by a body in only the nth second of motion, not the total displacement up to that time.
Why is the formula for displacement in the nth second useful?
It helps in solving kinematics problems efficiently, especially when finding how far an object moves in a particular second.
Can displacement in the nth second be negative?
Yes, if the object is moving in the negative direction (opposite to the initial direction of motion), then the displacement in the nth second can be negative.
Can the third equation of motion be used for non-uniform acceleration?
No, it is derived under the assumption of constant acceleration. If acceleration is non-uniform, the equation does not hold.
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Chapter Important Links
Learn more about related concepts such as motion, speed, velocity, displacement, equations of motion, scalar and vector quantities, and graphical representation of motion to understand acceleration more clearly. Similar concepts include uniform motion, non-uniform motion, average velocity, instantaneous velocity, and circular motion. Students should also study laws of motion, force, Newton’s laws, and kinematics problems to strengthen conceptual understanding and improve numerical-solving skills for CBSE, JEE, NEET, NDA, and other competitive examinations.