Derivation of the three equation s of motion

### Asked by mohini.ray | 5th Feb, 2020, 06:53: AM

###
(1) **First equation of Motion**: v = u + a t

Consider a body of mass m having initial velocity u.

Let after time t its final velocity becomes v due to uniform acceleration a.

Now we know that:

Acceleration = Change in velocity / Time taken

Acceleration = (Final velocity - Initial velocity) / Time taken

a = v - u / t

a t = v - u

or v = u + a t

This is the first equation of motion.

(2) **Second equation of motion**: s = u t + 1/2 at^{2}

Let the distance travelled by the body be s.

We know that

Distance = Average velocity x Time

Also, Average velocity = (u + v) / 2

.: Distance (t) = (u + v) / 2 t
.eq.(1)

Again we know that:

v = u + at

Substituting this value of v in eq.(1), we get

s = (u + u + a t) / 2 t

s = (2 u + a t) / 2 t

s = (2 u t + a t^{2}) / 2

s = (2 u t / 2) + (a t^{2 }/ 2)

or s = u t + (1/2) a t^{2}

This is the 2nd equation of motion.

(3) **Third equation of Motion**: v^{2} = u^{2} +2 a s

We know that

v = u + a t

v - u = a t

or t = (v - u) / a
..eq.(2)

Also we know that

Distance = average velocity x Time

.: s = [(v + u) / 2] x [(v - u) / a]

s = (v^{2} u^{2}) / 2 a

2 a s = v^{2} u^{2}

or v^{2} = u^{2} + 2 a s

This is the third equation of motion.

(1) **First equation of Motion**: v = u + a t

Consider a body of mass m having initial velocity u.

Let after time t its final velocity becomes v due to uniform acceleration a.

Now we know that:

Acceleration = Change in velocity / Time taken

Acceleration = (Final velocity - Initial velocity) / Time taken

a = v - u / t

a t = v - u

or v = u + a t

This is the first equation of motion.

(2) **Second equation of motion**: s = u t + 1/2 at^{2}

Let the distance travelled by the body be s.

We know that

Distance = Average velocity x Time

Also, Average velocity = (u + v) / 2

.: Distance (t) = (u + v) / 2 t .eq.(1)

Again we know that:

v = u + at

Substituting this value of v in eq.(1), we get

s = (u + u + a t) / 2 t

s = (2 u + a t) / 2 t

s = (2 u t + a t^{2}) / 2

s = (2 u t / 2) + (a t^{2 }/ 2)

or s = u t + (1/2) a t^{2}

This is the 2nd equation of motion.

(3) **Third equation of Motion**: v^{2} = u^{2} +2 a s

We know that

v = u + a t

v - u = a t

or t = (v - u) / a ..eq.(2)

Also we know that

Distance = average velocity x Time

.: s = [(v + u) / 2] x [(v - u) / a]

s = (v^{2} u^{2}) / 2 a

2 a s = v^{2} u^{2}

or v^{2} = u^{2} + 2 a s

This is the third equation of motion.

### Answered by Shiwani Sawant | 5th Feb, 2020, 10:08: AM

## Related Videos

- Derive an equation for position -time (S=ut+1/2at^2) relation for an object that travelled a distance 's' in time 't' under uniform acceleration 'a'
- a train is moving with an initial velocity of 30 metre per second the brakes are applied so as to produce a uniform acceleration of -1.5 metre per second square calculate the timeuniforma uniform acceleration
- an object is moving with a velocity 5m/s find the distance travelled in 5sec if the object is accelerating with a uniform acceleratrion of 5 m/s
- A body is thrown vertically upward with velocity u, the greatest height h to which it will rise is
- what is nth second of motion
- A car starts from the rest and accelerates at the rate of 5m/s2. It continue accelerating for 10 s and then move at constant velocity for next 10 s. Calculate the total distance traveled by the car in 20s. (5)
- A car is coasting backwards down a hill at -3.0 m/s when the driver gets the engine started. After 2.5s, the car is moving uphill at a velocity of +4.5 m/s. What is the car's acceleration?
- Problems and solutions on kinematics, Horizontal motion
- Please solve it with method

### Kindly Sign up for a personalised experience

- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions

#### Sign Up

#### Verify mobile number

Enter the OTP sent to your number

Change