Topic: Derivation of Equation of Linear Motion

Equations of Motion

The variable quantities in a uniformly accelerated rectilinear motion are time, speed, distance covered and acceleration. Simple relations exist between these quantities. These relations are expressed in terms of equations called equations of motion.

There are three equations of motion.

1) v = u+at

2) S = ut + 12 at^{2 }3) v^{2} = u^{2} + 2as

Where,

v = Final Velocity

u = Initial velocity

a = acceleration

s = distance traveled by a body

t = time taken.

Derivation of Equation of Motion

V = s/t

A = (v – u)/t

at = v-u

=> v-u = at

This is Newton’s First equation of motion. As you can you see, we can use this equation to calculate the velocity of a body which underwent an acceleration of a m/s for a time period of t seconds, provided we know the initial velocity of the body. Initial velocity i.e. u is the velocity of the body just before the body started to accelerate i.e. the velocity at t=0.

In case, the body started to accelerate from rest then we can substitute the value of initial velocity to be u = 0.

We sometimes also may want to find the total distance traveled by moving body.

A moving body might be either moving with a uniform velocity or with a uniform acceleration or even with a non-uniform acceleration.

In case of a body moving with a uniform velocity v, it is quite simple to calculate the total distance s traveled by the body in a time t. we know that velocity = distance traveled / time taken

v = s/t

=> s= vt

Thus, distance traveled = velocity x time

Now the situation is slightly different for a body moving with a uniform acceleration a. To calculate the distance traveled by an uniformly accelerating body, we derive the equation as follows.

If u is the initial velocity of an uniformly accelerating body and v is its velocity after a time t, then since the acceleration is UNIFORM, we can find the average velocity of the body as follows average velocity = (u+v)/2

Now, the distance s, traveled in the time t by the body is given by distance traveled = average velocity x time

s = [(u+v)/2]t

From equation (1) we have v = u+at, substituting this in the above equation for v, we get

s = [(u+u+at)/2]t

=> s = [(2u+at)/2]t

=> s = [(u + (1/2)at)]t

This is Newton’s second equation of Motion. This equation can be used to calculate the distance traveled by a body moving with a uniform acceleration in a time t. Again here, if the body started from rest, then we shall substitute u = 0 in this equation.

If you take a close look at the 2 equations of motion we derived just now you can observe that none of these equations carry a relation between distance traveled and final velocity of the body. All other relations are available. So, there is a need to find an equation which relates s and v. We derive it as follows.

We start with squaring equation (1). Thus we have

v^{2} = (u+at)^{2}

=> v^{2} = u^{2} + a^{2}t^{2} + 2uat

=> v^{2} = u^{2} + 2uat + a^{2}t^{2 }

=> v^{2} = u^{2} + 2a(ut + (1/2)at^{2})

now, using equation 2 we have

=> v^{2} = u^{2} + 2as – (3)

As you can see, the above equation gives a relation between the final velocity v of the body and the distance s traveled by the body.

Thus, we have the three Newton’s equations of Motion as:

(1) v= u + at

(2) s = ut + (1/2)at^{2}

(3) v^{2} = u^{2} + 2as

What is motion under gravity?

The attraction of the earth on the thrown object is called ‘Gravitational attraction’ or ‘gravity’, and the change in velocity of the object, as the result of this attraction, is known as ‘acceleration due to gravity’.

What Effect Does Gravity Have on Motion?

The pull of gravity between two objects depends on more than just the objects. Gravity’s effect also depends on the distance between the objects. Objects that are closer together have a greater attraction between them. The attraction is weaker when they are farther apart. Gravity exists wherever there is mass, such as in stars and planets. The gravity of each of these objects affects other objects in space. Earth’s gravity, for example, reaches millions and millions of kilometers into space. It grows weaker the farther away from Earth you get. Recall that because the moon is less massive than Earth, an astronaut standing on the moon weighs only one-sixth as much as on Earth. The astronaut’s weight would change between Earth and the moon. The effect of Earth’s gravity becomes less as the distance from Earth increases.

Gravitational Force

The gravitational force of attraction exists between objects. This force is the reason you feel yourself pulled towards the Earth.

How does the magnitude of the gravitational force between two objects change when the masses of the objects increase, and when their separation increases?

The gravitational force that you experience depends on the mass of the Earth, and on the separation between you and the mass of the Earth. For a large (approximately) spherical object like the Earth, it is the separation between you and the centre of the Earth that is relevant, and this separation is essentially the same for all objects close to the Earth’s surface.

Does every object on the Earth’s surface experience the same gravitational force?

In fact, the gravitational force *F*_{g} experienced by an object is *proportional* to its mass *m* (Figure 1), so we can write F_{g }µ M.

Thus the force of gravity you experience due to the Earth is proportional to your own mass. Someone with a mass of 50 kg experiences only half the gravitational force felt by someone with a mass of 100 kg.

Question

A body starting from rest and executing an accelerated motion covers a distance of 9 cm in 6 seconds. Calculate

1. The acceleration

A. 0.5 m/s^{2 }B. 1.0 m/s^{2} C. 1.7 m/s^{2} D. 2.0 m/s^{2}. Answer is A 0.5 m/s^{2}

2. The final velocity.

A. 4.0 m/s B. 3.0 m/s C. 7 m/s D. 8 m/s. Answer is B 3.0 m/s

A bullet is fired vertically with an initial velocity of 29.4 m/s

3. How high will it reach?

A. 34 m B. 54.7 m C. 30.9 m D. 44.1 m. Answer is D. 44.1 m

4. What is the time taken to reach that height?

A. 5 secs B. 3 secs C. 4.5 secs D. 9.5 secs. Answer is B 3 secs

5. A body of mass 2kg falls freely from rest through a height of 50 m and comes to rest having penetrated 5.0 cm of sand.

Calculate the velocity with which the body hits the sand.

A. 46 ms^{-1 }B. ^{ }76 ms^{-1 }C. 31.62 ms^{-1 }D. 45.53 ms^{-1}. Answer is C. 31.62 ms^{-1}

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