•  Acceleration\quad =\quad \frac { change\quad in\quad velocity }{ time\quad taken\quad for\quad change } \quad or\quad a\quad =\quad \frac { \Delta v }{ \Delta t }
  • Five constant acceleration equations:
 v\quad =\quad u\quad +\quad at   s\quad =\quad ut\quad +\quad \frac { 1 }{ 2 } a{ t }^{ 2 }  s\quad =\quad \frac { 1 }{ 2 } \left( u\quad +\quad v \right) t { v }^{ 2 }\quad =\quad { u }^{ 2\quad }+\quad 2as s\quad =\quad vt\quad -\quad \frac { 1 }{ 2 } a{ t }^{ 2 }
  • Acceleration of free falling object 9.8 { m }{ { s }^{ -2 } }

Revision of basic concept of Kinematics

To start off with we revise a few basic concepts involved in kinematics:

  1.  Distance: It is the length of a given path covered by a moving particle. It is a scaler quantity and it is measured in metre (m).
  2.  Displacement: It is the position of a particle from its starting point and it is a vector quantity. Displacement on the right of the reference point is considered positive while the displacement on the left of the reference point is considered negative.
  3.  Speed: It is the rate of change of distance. The speed shows the rate of motion with only magnitude and no direction. Hence, it is a scalar quantity. Speed is measured in metre per second m/s.
  4.  Velocity: It is defined as the rate of change of displacement. It is basically speed in a specified direction. It is also measured in metre per second. The symbol for speed is s, and the symbol for velocity is v. Be careful not to confuse the s for speed with s for displacement or s for seconds.
  5.  Acceleration: It is defined as the rate of change of velocity with respect to time. If a body is moving with uniform , steady or maximum velocity then its acceleration is zero. Acceleration is measured in  { m }/{ { s }^{ 2 } }  or { m }{ { s }^{ -2 } }.

To begin with if we say a particle is moving at a constant speed then we know there is no acceleration. Therefore we can simply use the formula speed\quad =\quad \frac { distance }{ time } to find the constant speed at which that particle is moving.

However, for a moving particle when its speed is varying with direction involved with respect to time it is said to be accelerating uniformly. When we say uniform or constant acceleration we assume there is no air resistance, no friction, and no other losses.

Constant Acceleration Equations

To work out acceleration there are five constant acceleration equations:

  1.   v\quad =\quad u\quad +\quad at 
  2.   s\quad =\quad ut\quad +\quad \frac { 1 }{ 2 } a{ t }^{ 2 }
  3.   s\quad =\quad \frac { 1 }{ 2 } \left( u\quad +\quad v \right) t
  4.  { v }^{ 2 }\quad =\quad { u }^{ 2\quad }+\quad 2as
  5.  s\quad =\quad vt\quad -\quad \frac { 1 }{ 2 } a{ t }^{ 2 }

These equations are for constant acceleration only.

The variables involved in the above equations are:

u\quad =\quad initial\quad speed\quad or\quad velocity\quad in\quad m{ s }^{ -1 } v\quad =\quad final\quad speed\quad or\quad velocity\quad in\quad m{ s }^{ -1 } a\quad =\quad uniform\quad acceleration\quad in\quad m{ s }^{ -2 } s\quad =\quad distance\quad travelled\quad or\quad displacement\quad in\quad m t\quad =\quad time\quad that\quad passes\quad in\quad seconds\quad \left( s \right)

Furthermore, note that when we have (velocity vs time graph) , for motion with uniform acceleration or deceleration the graph consists of straight lines. Also the gradient of these lines represent acceleration or deceleration. The area under the graph represents the distance travelled.

Note: In case of a free-falling object, it means its moving freely under gravity so acceleration is denoted as “g” . Mostly the free-falling object has an acceleration of approximately  9.81\quad { m }{ { s }^{ -2 } }. Hence value of g\quad \approx \quad 9.81\quad m{ s }^{ -2 }. g is known as acceleration for any motion under gravity.

Example #1

Q.  A car moves with constant acceleration along a straight level road. The car’s speed increases from 20\quad m{ s }^{ -1 } to 30\quad m{ s }^{ -1 } while it travels a distance of 200m.

i) The acceleration of the car
ii) The time taken to cover the distance


It is given to us that
initial\quad velocity\quad u=\quad 20m{ s }^{ -1 }
final\quad velocity\quad v=30\quad m{ s }^{ -1 }
Distance\quad s=\quad 200m

Looking at the information that we have from the question; to find a we use equation 4 from above equations.


i) { v }^{ 2 }\quad =\quad { u }^{ 2 }\quad +\quad 2as

{ (30) }^{ 2 }\quad =\quad { (20) }^{ 2 }\quad +\quad 2a(200) 900\quad -\quad 400\quad =\quad 400a 500\quad =\quad 400a a\quad =\quad \frac { 5 }{ 4 } \quad =\quad 1.25m{ s }^{ -2 }

ii) Now that we have the acceleration we can use equation 1 to find the time

v\quad =\quad u\quad +\quad at 30\quad =\quad 20\quad +\quad (1.25)t 10\quad =\quad 1.25t t\quad =\quad \frac { 10 }{ 1.25 } \quad =\quad 8\quad seconds


Example #2

Q.  A pebble is dropped into a well 18m deep and moves freely under gravity until it hits the bottom. Calculate the time it takes to reach the bottom. ( Take g\quad =\quad 9.8m{ s }^{ -2 })


From the question we know that
u=\quad 0\quad as\quad the\quad pebble\quad was\quad dropped
a\quad =g\quad =\quad 9.8m{ s }^{ -2 }
s\quad =\quad 18m

We substitute the above values in equation 2.


s\quad =\quad ut\quad +\quad \frac { 1 }{ 2 } a{ t }^{ 2 } 18\quad =\quad (0\quad \times \quad t)\quad +\quad \frac { 1 }{ 2 } \quad 9.8({ t }^{ 2 }) 18\quad =\quad 4.9{ t }^{ 2 }  { t }^{ 2 }\quad =\quad 3.67 t\quad =\quad \sqrt { 3.67 } \quad =\quad 1.92\quad s
CGP AS Level Mathematics Edexcel, Complete revision and practice