Linear Motion Equations: Detailed Derivation and Applications

Linear Motion is about bodies moving along a straight line even if it is through very short distance. Consider a body moving along a straight line accelerating uniformly from velocity u to final velocity v within time t. If we represent the distance covered between the initial and final velocity to be s; then there are three equations that can represent such a movement:

1. v=u + at

2. s=ut + (1/2)at²

3. v² = u²+2as

Deriving First equation

Multiplying by t on both sides:

at = v-u

Making v the subject by adding u on both sides:

at + u = v-u + u

Hence

v=u + at ————-(ii)

Deriving second equation of linear motion

From linear motions;

we can as well obtain velocities at different points in the motion and then divide by the number of points to get the average velocity.

Similarly given the initial velocity u and final velocity v, we can obtain average velocity as :

But

distance s= average velocity x time t , that is;

From the first equation;

v=u + at

Hence

and opening the bracket by multiplying by t on both sides, we get;

and therefore second equation is usually stated as:

Deriving third equation of linear motion.

From the equation; v=u + at

the equation below shows that time is the difference between the two velocities divided acceleration. making t the subject of the formula.

but also, average velocity is total displacement s divided by total time t. that is;

cross multiplying the equation above , we make distance s the subject of the formula so as we obtains:

2s=(v+u)t

and substituting t for v-u/a, we have;

2s=(v+u)(v-u)/2

and expanding the brackets we have

and so we have;

but -uv + uv = 0 and so we get

and so the equation above is reduced to the one below.

2as=v²-u²

and then rearranging the equation to get:

v²=u²+2as ———(iii)

Example problem

A body moving with uniform acceleration of 10ms^-1 covers a distance of 320m. If its initial velocity was 60ms^-1, calculate its final velocity.

solution

V² = u² +2as

form the equation:

v= 60ms^-1

distance s= 320m

a = 10ms^-1

v² = (60)² + 2 x 10 x 320

= 3600 + 6400 = 10000

v = sqrt(10000) = 100ms^-1

hence the final velocity is 100ms^-1

Exercise problems involving linear motion

1. A car accelerates from rest at a rate of 3 ms^-1. How long will it take for the car to reach a speed of 30m/s?
2. A train decelerates at a rate of 2m/s² until it comes to a complete stop. If the initial speed of the train is 25m/s, how far will it travel before stopping?
3. An object is thrown vertically upward with an initial velocity of 20 m/s. How long will it take for the object to reach its maximum height?
4. A ball is dropped from a height of 50 m. What is its velocity after 3s?
5. A cyclist travels at a constant speed of 10m/s for 20 s. How far does the cyclist travel during this time?
6. A stone is thrown horizontally from the top of a cliff with a velocity of 15m/s. How far does the stone travel horizontally before hitting the ground if it takes 3s to reach the ground?
7. An airplane accelerates down a runway at a rate of 2m/s² until it reaches a takeoff speed of 60m/s. If the runway is 1500 m long, how long does it take for the airplane to take off?
8. A rocket is launched vertically upward with an initial velocity of 50m/s. How high does the rocket go before it starts to fall back down?

Related Topics

• Introduction to forces
• Language of thin lenses