Derivation of Equation of Motion - Derivation First, Second & Third Equation of Motion

Derivation of Equation of Motion - 3 Equations of Motion

Edited By Vishal kumar | Updated on Jul 02, 2025 04:26 PM IST

Derivation of Equation of Motion - 3 equations of motion relate the displacement of an associated object with its rate, acceleration, and time. The motion of the associated object will follow many various ways. Here we are going to specialize in motion during a line (one dimension). we will so simply use positive and negative magnitudes of the displacement, rate, and acceleration, wherever negative values are within the other way to positive quantities.
If there's no acceleration, we have the acquainted formula
s=vt

Derivation of Equation of Motion - 3 Equations of Motion

where s is the displacement, v is the (constant) speed, and t is the time over which the motion occurred. This is simply a special case (a=0) of a lot of general derivation of equations of motion for constant acceleration below.

NCERT Physics Notes :

For a continuing acceleration a, initial speed u and an initial position of zero:

Variable	Equation
Velocity	v=u+at
Displacement(positive acceleration)	s = s₀ + ut + ½at²
Displacement(negative acceleration)	s = s₀ + ut - ½at²

The relation between speed and time could be a STRAIGHT one throughout the uniformly accelerated, straight-line motion. The longer the acceleration, the larger the modification in speed. change in speed is directly proportional to time once acceleration is constant. If speed is increased by a particular quantity during a certain time, it ought to increase by double that quantity in double the time. If an associate degree object already started with a particular speed, then its new speed would be the previous speed and this transformation. You must be compelled to be able to see the equation in your mind already. This is the best of the 3 equations to the derivation of equations of motion victimization pure mathematics. begin from the definition of acceleration.

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Derivation of First Equations of Motion:

$$
a=\Delta v / \Delta t
$$

Expand $\Delta v$ to $v-u$ and condense $\Delta t$ to $t$.

$$
a=(v-u) / t
$$
Then solve for $v$ as a function of $t$.

$$
\mathrm{v}=\mathrm{u}+\mathrm{at}
$$

Derivation of Second Equations of Motion:

Start with the definition of average velocity.

$$
v=\Delta s / \Delta t
$$

Expand $\Delta s$ to $s-s_0$ and condense $\Delta t$ to $t$.

$$
v=\left(s-s_0\right) / t
$$
Solve for the position.

$$
\begin{aligned}
& s=s_0+v t \ldots[a] \\
& v=1 / 2(v+u) \ldots \ldots[4]
\end{aligned}
$$
To continue, we'd like to resort to a touch trick referred to as the mean speed theorem or the Merton rule. I prefer the latter since the rule is applied to any amount that changes at an even rate — not simply speed. The Merton rule was 1st revealed in 1335 at Merton faculty, Oxford by land thinker, a man of science, logician, and calculator William Heytesbury (1313–1372). Once the speed of modification of an amount is constant, its average price is halfway between its final and initial values.

Substitute the first equations of motion [1] into equation [4] and simplify with the intent of eliminating v.

$$
\begin{aligned}
& v=1 / 2[(u+a t)+u] \text { Now substitute }[b] \text { into }[a] \text { to eliminate } v \\
& v=1 / 2(2 u+a t) \\
& v=u+1 / 2 a t \ldots[b]
\end{aligned}
$$

And finally, solve for s as a function of t .

$$
s=s_0+u t+1 / 2 a t^2 \ldots \ldots . .[2]
$$
This is the second equation of motion. It's written sort of a polynomial —

a relentless term $\left(\mathrm{s}_0\right)$, followed by a primary order term (ut ), followed by a second-order term ( $1 / 2 \mathrm{at}^2$ ).

The image $s_0$ is commonly thought of because of the initial position. The image $s$ is that the position it slows to later. you'll decide the ultimate position if you need to. The amendment in position ( $\Delta s$ ) is termed the displacement or distance (depending on circumstances) and a few individuals like writing the second equations of motion or derivation of equations of motion like this.

$$
\Delta s=u t+1 / 2 a t^2
$$

Related Topics,

Derivation of Third Equations of Motion:

The first 2 equations of the motion formula each describe one kinematic variable as an operation of your time. In essence…

1. Speed is directly proportional to time once acceleration is constant $(v \propto t)$.
2. Displacement is proportional to time square once acceleration is constant $\left(\Delta s \propto t^2\right)$.

3. Displacement is proportional to speed square once acceleration is constant ( $\Delta s \propto v^2$ ). This statement is especially relevant to driving safety. Once you double the speed of an automobile, it takes fourfold a lot of distance to prevent it. Triple the speed and you may want ninefold a lot of distance. This is often an honest rule of thumb to recollect. The abstract introduction is finished.Time to derive the formula derivation of equations of motion. method 1 Combine the primary 2 equations along in a very manner that may eliminate time as a variable. $v=v_0+$ at..... [1]

solve it for time…

$\mathrm{t}=\left(\mathrm{v}-\mathrm{v}_0\right) /$ aand then substitute it into the second equations of motion formula...

$$
s=s_0+v_0 t+1 / 2 a t^2
$$
Substitute the value of $t$ in equation [2] and solving we get

$$
\begin{aligned}
& 2 a\left(s-s_0\right)=v^2-v_0^2 \\
& v^2=v_0^2+2 a\left(s-s_0\right)
\end{aligned}
$$
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Frequently Asked Questions (FAQs)

1. A man is moving with 36 kmph. The time of reaction is 0.9 seconds. On seeing an obstacle in the path, he applies brakes and decelerates at 5 m/s , the total distance covered before he stops is: 1. 19m 2. 17m 3. 16m 4. 18m

19m

2. The ratio of the heights from which two bodies are dropped is 3:5 respectively. The ratio of their final velocities is: 1. 9:25 2. 3:5 3. 5:3 4. 5:3

3:5

3. If you whirl a stone on the end of the string and the string suddenly breaks, the stone will: 1. Fly off along the tangent to its circular path. 2. Fly directly towards you. 3. Spiral away from your hand. 4. Spiral in towards your hand.

Fly off along the tangent to its circular path.

4. The time taken by a train to slow down from 80 km/h to 20 km/h with a uniform deceleration of 2 m/s is: 1. 9s 2. 10s 3. 8.3s 4. 8.4s

8.3s

5. A body starts to slide over a horizontal surface with an initial velocity of 0.2 m/s. Due to friction, its velocity decreases at the rate of 0.02 m/s . How much time will it take for the body to stop? 1. 2.s 2. 15s 3. 10s 4. 5s

10s

6. How do the equations of motion change if the initial velocity is zero?

If the initial velocity (u) is zero, the equations of motion simplify to:

7. Why is it important to use consistent units when applying the equations of motion?

Using consistent units is crucial when applying the equations of motion to ensure accurate results. Each term in the equations must have the same units for the mathematical operations to be valid. For example, in v = u + at, velocity (v and u) must be in m/s, acceleration (a) in m/s², and time (t) in seconds. Mixing units (like using km/h for velocity and m/s² for acceleration) will lead to incorrect calculations.

8. How can we interpret the equation v = u + at in terms of a velocity-time graph?

In a velocity-time graph, the equation v = u + at represents a straight line. The initial velocity (u) is the y-intercept of this line, while the acceleration (a) is the slope of the line. The final velocity (v) at any time (t) can be read directly from the graph. This linear relationship on the v-t graph is a visual representation of constant acceleration.

9. Can the equations of motion be used for non-uniform acceleration?

The standard equations of motion (v = u + at, s = ut + (1/2)at², v² = u² + 2as) are derived assuming constant acceleration. They cannot be directly applied to situations with non-uniform acceleration. For non-uniform acceleration, calculus-based methods or numerical approximations are typically used to analyze the motion.

10. How can the equations of motion be used to solve problems involving free fall?

The equations of motion can be directly applied to free fall problems by using the acceleration due to gravity (g ≈ 9.8 m/s²) as the constant acceleration. Usually, we consider the downward direction as positive. For example, for an object dropped from rest:

11. What assumptions are made in deriving the equations of motion?

The key assumptions in deriving the equations of motion are:

12. What are the three equations of motion and why are they important?

The three equations of motion are:

13. How are the equations of motion derived?

The equations of motion are derived from the definitions of velocity and acceleration, assuming constant acceleration. Starting with the definition of acceleration (a = dv/dt), we integrate to get v = u + at. Then, using the definition of velocity (v = ds/dt), we integrate again to get s = ut + (1/2)at². The third equation is derived by eliminating time from the first two equations.

14. How does the equation v = u + at relate to the concept of average velocity?

The equation v = u + at is related to average velocity in that it represents a special case where acceleration is constant. In this case, the average velocity (v_avg) is simply the arithmetic mean of the initial and final velocities: v_avg = (u + v)/2. This relationship can be derived from v = u + at by substituting v and solving for the average velocity.

15. Why do we need three equations of motion? Isn't one enough?

We need three equations of motion because different problems provide different known variables. Each equation is useful for specific scenarios:

16. Can the equations of motion be used for objects moving in two or three dimensions?

The equations of motion as presented (v = u + at, s = ut + (1/2)at², v² = u² + 2as) are specifically for one-dimensional motion. However, they can be applied to two or three-dimensional motion by treating each dimension independently. For example, in projectile motion, we can use these equations separately for the horizontal and vertical components of motion.

17. Why does the third equation of motion (v² = u² + 2as) not involve time?

The equation v² = u² + 2as is derived by eliminating time from the other two equations of motion. This is useful in situations where we don't know or don't need to know the time taken for the motion, but we have information about displacement, initial velocity, and final velocity. It allows us to solve problems directly without having to calculate time as an intermediate step.

18. What is the significance of the negative sign in the equations of motion when dealing with deceleration?

The negative sign in the equations of motion when dealing with deceleration indicates that the acceleration is in the opposite direction to the velocity. For example, in v = u + at, if a is negative, it means the object is slowing down. The negative sign ensures that the equations correctly represent the physics of the situation, showing that the velocity is decreasing over time.

19. How does air resistance affect the applicability of the equations of motion?

The standard equations of motion assume negligible air resistance. In reality, air resistance can significantly affect motion, especially at higher speeds or for longer durations. Air resistance typically opposes motion and increases with velocity, leading to non-constant acceleration. For motions where air resistance is significant, more complex models or numerical methods are needed to accurately describe the motion.

20. Can the equations of motion be applied to circular motion?

The standard equations of motion (v = u + at, s = ut + (1/2)at², v² = u² + 2as) are for linear motion and cannot be directly applied to circular motion. In circular motion, the direction of velocity is constantly changing, even if speed is constant. For uniform circular motion, we use different equations involving angular velocity and centripetal acceleration. However, the concepts behind the equations of motion can be adapted for rotational motion using angular analogues.

21. How do the equations of motion relate to the concept of relative motion?

The equations of motion can be applied to relative motion by considering the relative velocities and accelerations between objects or reference frames. For example, if we have two objects moving along the same line, we can apply the equations to their relative motion by using their relative initial velocity and acceleration. The displacement calculated would then be the change in distance between the objects.

22. What is the significance of the equation v² = u² + 2as in analyzing stopping distances?

The equation v² = u² + 2as is particularly useful in analyzing stopping distances because it doesn't involve time explicitly. In stopping distance problems, we often know the initial velocity (u) and final velocity (v = 0), and we want to find the distance (s) needed to stop given a certain deceleration (a). This equation allows us to calculate the stopping distance directly without needing to know the time taken to stop.

23. How can the equations of motion be used to analyze the motion of a car during emergency braking?

During emergency braking, we can use the equations of motion, particularly v² = u² + 2as, to analyze the car's motion. Here, u is the initial speed, v is the final speed (usually zero), a is the deceleration (negative), and s is the stopping distance. By knowing the coefficient of friction between the tires and road, we can determine the maximum deceleration and then calculate the minimum stopping distance or the time taken to stop.

24. What is the physical meaning of the intersection point of displacement-time graphs for two objects, and how can it be found using equations of motion?

The intersection point of displacement-time graphs for two objects represents the time and position where the objects meet or pass each other. To find this point using equations of motion:

25. How do the equations of motion help in understanding the concept of terminal velocity?

The equations of motion, particularly v = u + at, help us understand terminal velocity conceptually. In free fall with air resistance, as velocity increases, air resistance increases until it balances the gravitational force. At this point, acceleration becomes zero, and the object reaches terminal velocity. While the standard equations don't directly model this (as acceleration is not constant with air resistance), they help us understand that terminal velocity is reached when a = 0 in v = u + at, meaning velocity stops changing.

26. What is the relationship between the equations of motion and the concept of jerk (rate of change of acceleration)?

The standard equations of motion assume constant acceleration, which means the jerk (rate of change of acceleration) is zero. In more complex motions where jerk is non-zero, the acceleration changes over time, and the standard equations of motion no longer apply directly. Understanding jerk helps in recognizing the limitations of these equations and when more advanced mathematical tools are needed to describe motion accurately.

27. What is the significance of the equation s = vt - (1/2)at² in analyzing the motion of an object coming to a stop?

The equation s = vt - (1/2)at² is useful for analyzing an object coming to a stop because it relates the stopping distance (s) to the initial velocity (v), time taken to stop (t), and deceleration (a). The negative sign before (1/2)at² indicates that the acceleration is in the opposite direction to the initial motion. This equation helps in calculating stopping distances or times for vehicles, which is crucial for road safety and design.

28. How does the equation s = ut + (1/2)at² represent the displacement graphically?

The equation s = ut + (1/2)at² represents a parabola when graphed with displacement (s) on the y-axis and time (t) on the x-axis. The ut term represents the initial linear component of the motion, while the (1/2)at² term adds the curvature due to acceleration. For constant positive acceleration, the parabola opens upward; for constant negative acceleration, it opens downward.

29. What does the term (1/2)at² in s = ut + (1/2)at² represent physically?

The term (1/2)at² in the equation s = ut + (1/2)at² represents the additional displacement due to acceleration. Physically, it accounts for the fact that as an object accelerates, it covers more distance than it would if it were moving at a constant velocity. The factor of 1/2 arises from the averaging of the increasing velocity over time.

30. How does the equation v² = u² + 2as relate to the work-energy theorem?

The equation v² = u² + 2as is closely related to the work-energy theorem. If we multiply both sides by m/2 (where m is mass), we get:

31. How do the equations of motion relate to the concept of instantaneous velocity?

The equations of motion deal with average velocities over time intervals, but they can approximate instantaneous velocity in the limit of small time intervals. The equation v = u + at gives the instantaneous velocity at time t, assuming constant acceleration. In calculus terms, this equation is equivalent to integrating the instantaneous acceleration over time.

32. Can the equations of motion be used to predict the time and position where two moving objects will meet?

Yes, the equations of motion can be used to predict when and where two moving objects will meet. This typically involves setting up equations for each object's motion and then solving them simultaneously. For example, if two cars are moving towards each other, we can use s = ut + (1/2)at² for each car and set the total distance equal to the sum of their displacements to find the meeting time and position.

33. What is the physical interpretation of the area under a velocity-time graph, and how does it relate to the equations of motion?

The area under a velocity-time graph represents the displacement of the object. This concept is directly related to the equation s = ut + (1/2)at². For constant acceleration, the v-t graph is a straight line, and the area under this line is a trapezoid. The area of this trapezoid gives the displacement, which is exactly what the equation s = ut + (1/2)at² calculates.

34. How can the equations of motion be used to analyze the motion of an object thrown vertically upward?

For an object thrown vertically upward, we can use the equations of motion with a negative acceleration (due to gravity pulling downward). If we choose upward as positive:

35. What is the relationship between the three equations of motion and the concept of average acceleration?

The three equations of motion are derived assuming constant acceleration, which means the average acceleration is equal to the instantaneous acceleration at all times. The equation a = (v - u)/t, which defines average acceleration, is essentially a rearrangement of v = u + at. This relationship highlights that for constant acceleration, the change in velocity is directly proportional to the time elapsed.

36. How do the equations of motion change when considering motion on an inclined plane?

When considering motion on an inclined plane, the equations of motion remain the same in form, but the acceleration used is the component of gravity parallel to the incline. If θ is the angle of inclination:

37. How can the equations of motion be used to analyze the motion of a projectile, considering both horizontal and vertical components?

For projectile motion, we apply the equations of motion separately to the horizontal and vertical components:

38. How do the equations of motion relate to the concept of momentum and impulse?

While the equations of motion don't directly involve momentum, they are closely related. The equation v = u + at can be rewritten as m(v - u) = mat, where m is mass. The left side represents change in momentum, and the right side represents impulse (force × time). This shows that the equations of motion are consistent with the impulse-momentum theorem, linking kinematics (motion) to dynamics (forces and momentum).

39. Can the equations of motion be applied to objects moving with non-zero initial velocity in free fall?

Yes, the equations of motion can be applied to objects with non-zero initial velocity in free fall. We use: