# Derivation of Equation of Motion - 3 Equations of Motion

**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've the acquainted formula

s=vt

where s is that the displacement, v the (constant) speed and t the time over that 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.

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

Variable | Equation |

Velocity | v=u+at |

Displacement(positive acceleration) | s = s_{0} + ut + ½at^{2} |

Displacement(negative acceleration) | s = s_{0} + ut - ½at^{2} |

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=∆v/∆t

Expand ∆v to v − u and condense ∆t to t.

a=(v-u)/t

Then solve for v as a function of t.

v = u + at

## Derivation of Second Equations of Motion:

Start with the definition of average velocity.

v=∆s/∆t

Expand ∆s to s − s_{0} and condense ∆t to t.

v=(s-s_{0})/t

Solve for position.

s = s_{0} + vt ...[a]

v = ½(v + u)...... [4]

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

v = ½(2u+ at)

v = u + ½at.... [b]

Now substitute [b] into [a] to eliminate v

s = s_{0 }+ (u + ½at)t

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

s = s_{0} + ut + ½at^{2} .......[2]

**This is the second equation of motion. It's written sort of a polynomial — **

a relentless term (s_{0}), followed by a primary order term (ut ), followed by a second-order term (½at^{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 (∆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.

∆s = ut + ½at^{2}

## 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 ∝ t).

2. Displacement is proportional to time square once acceleration is constant (∆s ∝ t^{2}).

3. Displacement is proportional to speed square once acceleration is constant (∆s ∝ 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…

_{0})/a

and then substitute it into the second equations of motion formula…

s = s_{0} + v_{0}t + ½at^{2}......... [2]

Substitute the value of t in equation [2] and solving we get

2a(s-s_{0})=v^{2}-v_{0}^{2}

v^{2}=v_{0}^{2}+2a(s-s_{0})

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## Frequently Asked Question (FAQs) - Derivation of Equation of Motion - 3 Equations of Motion

**Question: **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

**Answer: **

19m

**Question: **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

**Answer: **

3:5

**Question: **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.

**Answer: **

Fly off along the tangent to its circular path.

**Question: **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

**Answer: **

8.3s

**Question: **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

**Answer: **

10s