Third Equation of Motion – Derivation and Problems

Introduction:

In a uniformly accelerated motion, the velocity of the body increases or decreases by equal amounts in equal intervals of time. For such kinds of motion, the various attributes of motion such as the displacement, velocity, and acceleration for a certain time interval can be related by a set of equations. This set of equations is called the equations of motion. One or two unknown quantities can be calculated by using the equations of motion if other quantities are known.

Explanation:

Equations of Motion:

Let us consider the following physical quantities by the symbols given below for a body in a uniformly accelerated or decelerated motion.

• Initial velocity = u
• Final velocity = v
• Acceleration = a
• Time = t
• Displacement = s

The equations of motion connecting the above-mentioned physical quantities are as follows:

v = u + at

s = ut + ½ at²

v² = u² + 2as

Derivation of the Third Equation of Motion:

All the equations of motion can be derived from the V-t graph. Consider the V-t graph of a body in a uniformly accelerated motion shown below. The v-t graph (AC) represents the motion of a body in a uniformly accelerated motion with a non-zero initial velocity. The constructions CD (perpendicular to the x-axis) and AB (perpendicular to CD) are drawn. Now, OACD forms a trapezium which is made of a triangle ACB and a rectangle OABD.

The above v-t graph represents the motion of a body with a non-zero initial velocity u (given by the length OA of the graph). The body increases its velocity to v (given by the length CD) in time t (given by the length OD). Let u be the displacement of the body in time t and the velocity changes at a uniform rate a.

Third Equation of Motion:

From the graph, it is seen that,

Displacements = Area (trapezium OACD)

                        =  

                        =

                      s =

                      2s = t (u + v) ————-(1)

From the first equation of motion, we have,

                       v = u + at

                 Or, t = (v – u)/a ————-(2)

On substituting equation (2) in equation (1) we get,

                     2s = (v + u) (v – u)/a

2as = v² – u²

v² = u² + 2as

Problems:

1. A body moving at a speed of 72 km/h accelerates uniformly at the rate of 2 m/s to cover 1.8 km. Calculate the final velocity of the body. How long did it take for the body to cover that distance?

Solution:

Given that,

The initial velocity, u = 72 km/h = 72 x (5/18) = 20 m/s

The acceleration, a = 2 m/s²

The distance covered, s = 1.8 km = 1800 m

On using the third equation of motion we get,

       v² = u² + 2as

Or, v² = (20)² + 2 x 2 x 1800

Or, v² = 400 + 7200

Or, v² = 7600

 Or, v = 87.17 m/s

Therefore, the final velocity of the car is 87.17 m/s.

2. A car moving at a velocity of 5 m/s accelerates to 25 m/s while covering a distance of 1.8 km. Work out the acceleration of the car and the time it takes to cover that distance.

Solution:

Given that,

The initial velocity, u = 5 m/s

The final velocity, v = 25 m/s

The distance covered, s = 1.8 km = 1800 m

On using the third equation of motion we get,

      v² = u² + 2as

Or, (25)² = (5)² + 2 x a x 1800

Or, 625 = 25 + 3600a

Or, 3600 a = 625 – 25

Or, 3600 a = 600

Or, a = 600/ 3600 = 1.67 m/s²

Therefore, the acceleration of the car is 1.67 m/s².

On using the first equation of motion we get,

      v = u + at

Or, 25 = 5 + (1/6) x t

Or, t/6 = 20

Or, t = 20 x 6

Or, t = 120 s

Therefore, the time taken by the car is 120 s.

3. A bus starts from rest and attains a velocity of 90 km/h in 10 minutes. Find out the acceleration and the distance traveled by the bus to attain this velocity.

Solution:

Given that,

The initial velocity, u = 0

The final velocity, v = 90 km/h = 90 x (5/18) = 25 m/s

Time taken, t = 10 minutes = 60 x 10 = 600 s

On using the first equation of motion we get,

      v = u + at

Or, 25 = 0 + a x 600

Or, a = 25/600

Or, a = 1/24 m/s²

Therefore, the acceleration is 1/24 m/s².

On using the third equation of motion we get,

      v² = u² + 2as

Or, (25)² = 0 + 2 x (1/24) x s

Or, 625 = s/12

Or, s = 625 x 12

Or, s = 7500 m

Or, s = 7.5 km

Therefore, the distance covered by the bus is 7.5 km.

4. A ball is dropped from the top of a building of a height of 50 m. Considering the acceleration due to the gravity of the Earth to be 10 m/s², find out the velocity of the ball right before it touches the ground and the time it takes to hit the ground.

Solution:

Given that,

The initial velocity, u = 0

The acceleration, a = 10 m/s²

The displacement, s = 80 m

On using the third equation of motion we get,

      v² = u² + 2as

Or, (v)² = 0 + 2 x 10 x 80

Or, (v)² = 1600

Or, v = sqrt (1600)

Or, v = 40 m/s

Therefore, the velocity of the ball before hitting the ground is 40 m/s.

On using the first equation of motion we get,

      v = u + at

Or, 40 = 0 + 10 x t

Or, t = 40/10

Or, t = 4 s