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#### Newton’s Equation of Motion:

**First Equation of Motion:**

Let u = initial velocity of a body, v = final velocity of the body

t = time in which the change in velocity takes place.

By the definition of acceleration

Considering magnitudes only

v = u + at

This equation is known as Newton’s First equation of motion.

**Second Equation of Motion:**

Let u = initial velocity of a body, v = final velocity of the body

t = time in which the change in velocity takes place, a = acceleration of the body

This equation is known as Newton’s second equation of motion.

**Third Equation of Motion:**

Let u = initial velocity of a body, v = final velocity of the body

t = time in which the change in velocity takes place.

from equation (1) and (2)

Considering the magnitude only

v² = u² = 2 a s

This equation is known as Newton’s third equation of motion.

**Expression for the Distance Travelled by Body in nth Second of its Motion:**

By Newton’s Second equation of motion, s = ut + ½ at²

where s = displacement of b the dy in ‘t’ seconds

u = initial velocity of the body, a = acceleration of the body, t = time

The distance travelled by body in ‘n’ seconds is given by

This distance by travelled by the body in (n-1) seconds is given by

∴ The distance travelled by the body in nth second

**Motion Under Gravity:**

- A special case of uniform acceleration is the motion of a body under gravity. It is found that close to the surface of the earth, and in the absence of air resistance, all the bodies fall to the earth, at a given place, with the constant acceleration. This constant acceleration is called acceleration due to gravity or gravitational acceleration.
- It is denoted by “g”. It is always directed downward. Its magnitude is approximately 9.8 m/s
^{2}. - For motion of body under gravity Newton’s equations of motion can be written as

**Sign Convention (Cartesian):**

- All vectors directed towards the right of the reference point are considered positive.
- All vectors directed towards the left of the reference point are considered negative.
- All vectors directed vertically upward the reference point are considered positive.
- All vectors directed vertically downward the reference point are considered negative.
- By this sign convention acceleration due to gravity “g” is always negative.

#### Example – 1:

- A car acquires a velocity of 72 kmph in 10 s starting from rest. Calculate its average velocity, acceleration and distance travelled during this period.
**Solution:****Given:**Initial velocity = u = 0, Final velocity = v = 72 kmph = 72 x 5/18 = 20 m/s, Time taken = t = 10 s**To Find:**average velocity = v_{av}=?, acceleration = a =?, distance travelled = s =?

v_{av} = (u + v) /2 = (0 = 20)/2 = 10 m/s

Average velocity is 10 m/s

By first equation of motion we have

v = u + at

∴ 20 = 0 + a x 10

∴ a = 20/10 = 2 m/s^{2}

acceleration = 2 m/s^{2}

By second equation of motion

s = ut + ½ at^{2} = 0 x (10) + ½ x 2 x (10)^{2} = 0 + 100 = 100 m

distance travelled = 100 m

#### Example – 2:

- A ball is moving with a velocity of 0.5 m/s. Its velocity is decreasing at the rate of 0.05 m/s
^{2}. What is its velocity after 5 s? How much time will it take from start to stop? What is the distance travelled by it before stopping? **Solution:****Given:**Initial velocity = u = 0.5 m/s, acceleration = a = – 0.05 m/s^{2}**To Find:**a) v = ? when t = 5 s b) t = ? when v = 0, c) distance travelled = s = ?

By first equation of motion (v = ? when t = 5)

v = u + at = 0.5 + (- 0.05) x 5 = 0.5 – 0.25 = 0.25 m/s

Thus the velocity of ball after 5 s is 0.25 m/s

By First equation of motion (t = ? when v = 0)

v = u + at

∴ 0 = 0.5 + (- 0.05) x t

∴ 0.5 = – 0.05 x t

∴ t = 0.5/0.05 = 10 s

The ball will stop after 10 s from start

By the second equation of motion

s = ut + ½ at^{2} = 0.5 x (10) + ½ x (-0.05) x (10)^{2} = 5 – 2.5 = 2.5 m

The ball travels 2.5 m before coming to rest.

#### Example – 3:

- A car initially at rest starts moving with acceleration 0.5 m/s
^{2}covers a distance of 25 m. Calculate the time required to cover this distance and the final velocity of the car. **Solution:****Given:**Initial velocity = u = 0 m/s, acceleration = a = 0.5 m/s^{2}, distance travelled = s = 25m.

**To Find:**time taken = t =? , final velocity = v = ?

By the second equation of motion

s = ut + ½ at^{2}

∴ 25 = 0 x (t) + ½ x (0.5) x t^{2}

∴ 25 = 0.25 t^{2}

∴ t^{2 }= 25/ 0.25 = 100

∴ t = 10 s

Time required by car to cover distance of 25 m is 10 s

By first equation of motion

v = u + at = 0 + ( 0.5) x 10 = 5 m/s

The final velocity of ball after is 5 m/s.

#### Example – 4:

- A body starts from rest with a uniform acceleration of 2 m/s
^{2}. Calculate the distance travelled by the body in 2 s. **Solution:****Given:**Initial velocity = u = 0 m/s, acceleration = a = 2 m/s^{2}, time taken = t = 2 s.**To Find:**Distance travelled = s =?

By the second equation of motion

s = ut + ½ at^{2} = 0 x (2) + ½ x (2) x 2^{2 } = 4 m

The distance travelled by the body is 4 m

#### Example – 5:

- A body is moving with 5m/s is accelerated at 5 m/s
^{2}. Calculate the distance travelled by the body in 5 s. **Solution:****Given:**Initial velocity = u = 10 m/s, acceleration = a = 5 m/s^{2}, time taken = t = 5 s.

**To Find:**Distance travelled = s =?

By the second equation of motion

s = ut + ½ at^{2} = 10 x (5) + ½ x (5) x 5^{2 } = 50 + 62.5= 112.5 m

The distance travelled by the body is 4 m

#### Example – 6:

- A vehicle is moving at a velocity of 30 kmph at some instant. After 2 s its velocity is found to be 33.6 kmph and after further 2 s the velocity is found to be 37.2 kmph. find the acceleration of the vehicle and comment on the result.
**Solution:**- Consider the change of velocity from 30 kmph to 33.6 kmph in 2 s.

u = 30 kmph = 30 x 5/18 = 150/18 m/s, v = 33.6 kmph = 33.6 x 5/18 = 168/18 m/s, t = 2 s

By first equation of motion

v = u + at

168/18 = 150/18 + a(2)

168/18 – 150/18 = 2a

2a = 18/18 = 1

a = 1/2 = 0.5 m/s^{2}

- Consider the change of velocity from 33.6 kmph to 37.2 kmph in 2 s.

u = 33.6 x 5/18 = 168/18 m/s, v = 37.2 kmph = 37.2 x 5/18 = 186/18 m/s, , t = 2 s

By first equation of motion

v = u + at

186/18 = 168/18 + a(2)

186/18 – 168/18 = 2a

2a = 18/18 = 1

a = 1/2 = 0.5 m/s^{2}

The acceleration is same in both the cases thus the vehicle is moving with a constant acceleration of 0.5 m/s^{2}.

#### Example – 7:

- A body initially at rest travels a distance of 100 m in 5 s with constant acceleration. calculate the acceleration and the final velocity at the end of 5 s.
**Solution:****Given:**Initial velocity = u = 0 m/s, time taken = t = 5 s, distance travelled = s = 100 m**To Find:**Acceleration = a =?, final velocity = v =?

By the second equation of motion

s = ut + ½ at^{2}

100 = 0 x (5) + ½ x (a) x 5^{2 }

200 = 25 a

a = 200/25 = 8 m/s^{2}

Acceleration = 8 m/s^{2}

By first equation of motion

v = u + at

v = 0 + 8 x 5 = 40 m/s

Final velocity = 40 m/s

#### Example – 8:

- A body initially at moving with a speed of 18 kmph is accelerated uniformly at the rate of 9 cm/s
^{2}covers a distance of 200 m. Calculate the final velocity. **Solution:****Given:**Initial velocity = u = 18 kmph = 18 x 5/18 = 5 m/s, distance travelled = s = 100 m, acceleration = 9 cm/s^{2}= 0.09 m/s^{2}.**To Find:**final velocity = v =?

By the third equation of motion

v^{2} = u^{2} + 2as

v^{2} = 5^{2} + 2 x 0.09 x 200 = 25 + 36

v^{2} = 61

v = 7.81 m/s

Final velocity = 7.81 m/s

#### Example – 9:

- A body initially at rest starts accelerating at the rate of 2 m/s
^{2}. Find the final velocity and distance travelled by the body after 5s. **Solution:****Given:**Initial velocity = u = 0 m/s, acceleartion = a = 2 m/s^{2}, time taken = t = 5 s**To Find:**distance travelled = ?, final velocity = v =?

By the second equation of motion

s = ut + ½ at^{2} = 0 x 5 + ½ x 2 x 5^{2} = 25 m

Distance travelled = 25 m

By first equation of motion

v = u + at

v = 0 + 2 x 5 = 10 m/s

Final velocity = 10 m/s

#### Example – 10:

- A train moving with a velocity 72 kmph is brought to rest by applying brakes in 5 s. Calculate the retardation and distance travelled during this period.
**Solution:****Given:**Initial velocity = u = 72 kmph = 72 x 5/18 = 20 m/s, final velocity = v = 0 m/s, time taken = t = 5 s**To Find:**distance travelled = ?, retardation = a =?

By first equation of motion

v = u + at

0 = 20 + a x 5

5a = -20

a = -20/5 = -4 m/s^{2}

Neagative sign indicates retardation

retardation = 4 m/s^{2}

By the second equation of motion

s = ut + ½ at^{2} = 20 x 5 + ½ x (- 4) x 5^{2} = 100 – 50 = 50 m

Distance travelled = 50 m

#### Example – 11:

- A body moves from rest with uniform acceleration and travels 270 m in 3 s. Find the velocity of the body after 5 s..
**Solution:****Given:**Initial velocity = u = 0 m/s, distance travelled = s = 270 m, time taken = t = 3 s**To Find:**velocity = ? when t = 5 s.

By the second equation of motion

s = ut + ½ at^{2}

270 = 0 x 5 + ½ x (a) x 3^{2}

270 = 9/2 a

a = 540/9 = 60 m/s^{2}

By first equation of motion

v = u + at = 0 + 60 x 5 = 300 m/s

velocity after 5s is 300 m/s

#### Example – 12:

- A body moving with constant acceleration travels the distances 3 m and 8 m in 1 s and 2 s respectively. Calculate the initial velocity and acceleration of the body
**Solution:****Given:**Case – 1: distance travelled = s_{1}= 3 m, time taken = t_{1}= 1 s, Case – 2: distance travelled = s_{2}= 8 m, time taken = t_{2}= 2 s,**To Find:**initial velocity = u = ? acceleration = a =?

By the second equation of motion

s = ut + ½ at^{2}

For first case

3 = u x 1 + ½ x (a) x 1^{2}

3 = u + ½ x (a)

2u + a = 6 ……. (1)

For second case

8 = u x 2 + ½ x (a) x 2^{2}

8 = 2u + 2 a

u + a = 4 ……. (2)

Solving equations (1) and (2) simultaneously

a = 2 and u = 2

In itial velocity = 2 m/s and acceleration = 2 m/s^{2}.

#### Example – 13:

- A body moving with constant acceleration travels the distances 84 m and 264 m in 6 s and 11 s respectively. Calculate the initial velocity and acceleration of the body
**Solution:****Given:**Case – 1: distance travelled = s_{1}= 84 m, time taken = t_{1}= 6 s, Case – 2: distance travelled = s_{2}= 264 m, time taken = t_{2}= 11 s,**To Find:**initial velocity = u = ? acceleration = a =?

By the second equation of motion

s = ut + ½ at^{2}

For first case

84 = u x 6 + ½ x (a) x 6^{2}

84 = 6u + 18a

u + 3a = 14 ……. (1)

For second case

264 = u x 11 + ½ x (a) x 11^{2}

264 = 11u + ½ x (a) x 121

22u + 121a = 528

2u + 11a = 48 ……. (2)

Solving equations (1) and (2) simultaneously

a = 4 and u = 2

In itial velocity = 2 m/s and acceleration = 4 m/s^{2}.

#### Example – 14:

- A bus is moving with uniform velocity. The driver of the bus sees a pedestrian crossing the road at a distance of 60 m from the bus. He applied the brakes and reduce the speed with retardation of 25 cm/s2 and takes 20 s to stop the bus. Find the initial velocity of the bus and also decide the fate of the pedestrian.
**Solution:****Given:**acceleration = – 25 cm/s^{2}= – 0.25 m/s^{2}, time taken = t = 20 s, Final velocity = 0 m/s.**To Find:**initial velocity = u = ? Distance travelled = s = ?

By first equation of motion

v = u + at

0 = u + (-0.25)(20)

0 = u – 5

u = 5 m/s

Initial velocity = u = 5 m/s

By the second equation of motion

s = ut + ½ at^{2}

s = 5 x 20 + ½ x (-0.25) x 20^{2} = 100 – 50 = 50 m

As the distance travelled by the bus is less than the distance of pedestrian from the bus

There will be no accident

#### Example – 15:

- A train is moving with a velocity of 90 kmph. When the brakes are applied the acceleration is found to ne 0.5 m/s
^{2}. Find the velocity after 10 s, the time taken to stop and the distance traveled before stopping from the application of brakes. **Solution:****Given:**acceleration = – 0.5 m/s^{2}, time taken = t = 20 s, initial velocity = u = 90 kmph = 90 x 5/18 = 25 m/s.**To Find:**velocity = v = ? when t = 10 s, time taken = ? when v = 0, Distance travelled = s = ?

By first equation of motion

v = u + at = 25 + (-0.5)(10) = 25 – 5 = 20 m/s

Velocity after 10 s is 20 m/s

By first equation of motion

v = u + at

0 = 25 + (-0.5)(t)

– 25 = – 0.5 t

t = 25/0.5 = 50 s

The train will take 50 s to stop

Velocity after 10 s is 20 m/s

By the second equation of motion

s = ut + ½ at^{2}

s = 25 x 50 + ½ x (-0.5) x 50^{2} = 1250 – 625 = 625 m

The train will civer 625 m before coming to rest

#### Example – 16:

- A train is moving with a velocity of 54 kmph. It is accelerated at the rate 5 m/s
^{2}. Find the distance travelled by the train in 5 seconds and in the 5th second of its journey. **Solution:****Given:**acceleration = 5 m/s^{2}, time = t = 5 s, initial velocity = u = 54 kmph = 54 x 5/18 = 15 m/s.**To Find:**Distance travelled in 5 s =? and distance travelled in 5th second = ?

By the second equation of motion

s = ut + ½ at^{2}

s = 15 x 5 + ½ x (5) x 5^{2} = 75 + 62.5 = 137.5 m

Distance travelled in 5 s is 137.5 m

The distance travelled in nth second is given by

s_{n} = u +1/2a(2n – 1)

s_{5} = 15 +1/2x 5 x (2 x 5 – 1) = 15 +2.5 x 9

s_{5} = 15 + 22.5 = 37.5 m

The distance travelled in 5th second is 37.5 m

Science > Physics > Motion in Straight Line > You are Here |

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