- Introduction
- |
- Speed and Velocity
- |
- acceleration
- |
- Equations of uniformly accelerated motion
- |
- Graphical representation of motion
- |
- Equations of motion by graphical method
- |
- Uniform circular motion

- There are three equations of bodies moving with uniform acceleration which we can use to solve problems of motion

- The first equation of motion is \(v = u + at\) , where v is the final velocity and u is the initial velocity of the body.
- First equation of motion gives velocity acquired by body at any time \(t\).
- Now we know that acceleration

so, \(a = \frac{{v - u}}{t}\)

and, \(at = v - u\)

rearranging above equation we get first equation of motion that is

\(v = u + at\)

- Second equation of motion is

\(s = ut + \frac{1}{2}a{t^2}\)

where \(u\) is initial velocity, \(a\) is uniform acceleration and \(s\) is the distance travelled by body in time \(t\). - Second equation of motion gives distance travelled by a moving body in time \(t\).
- To obtain second equation of motion consider a body with initial velocity \(u\) moving with acceleration a for time \(t\) its final velocity at this time be \(v\). If body covered distance \(s\) in this time \(t\) , then average velocity of the body would be

Distance travelled by the body is

From first equation of motion

\(v = u + at\)

So putting first equation of motion in above equation we get ,

\(s = \frac{{u + u + at}}{2} \times t = \frac{{\left( {2u + at} \right)t}}{2} = \frac{{2ut + a{t^2}}}{2}\)

Rearranging it we get

\(s = ut + \frac{1}{2}a{t^2}\)

- Third equation of motion is

\({v^2} = {u^2} + 2as\) where \(u\) is initial velocity, \(v\) is the final velocity, \(a\) is uniform acceleration and \(s\) is the distance travelled by the body. - This equation gives the velocity acquired by the body in travelling a distance \(s\).
- Third equation of motion can be obtained by eliminating time t between first and second equations of motion.

So, first and second equations of motion respectively are

\(v = u + at\) and \(s = ut + \frac{1}{2}a{t^2}\)

Rearranging first equation of motion to find time t we get

\(t = \frac{{v - u}}{a}\)

Putting this value of t in second equation of motion we get

\(s = \frac{{u\left( {v - u} \right)}}{a} + \frac{1}{2}a{\left( {\frac{{v - u}}{a}} \right)^2}\)

\(s = \frac{{uv - {u^2}}}{a} + \frac{{a\left( {{v^2} + {u^2} - 2uv} \right)}}{{2{a^2}}}\)

\(s = \frac{{2uv - 2{u^2} + {v^2} + {u^2} - 2uv}}{{2a}}\)

Rearranging it we get

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

- These three equations of motion are used to solve uniformly accelerated motion problems and following three important points should be remembered while solving problems
- if a body starts moving from rest its initial velocity \(u = 0\)
- if a body comes to rest i.e., it stops then its final velocity would be \(v = 0\)
- If a body moves with uniform velocity then its acceleration would be zero.

Class 9 Maths Home page Class 9 Science Home page

- Biology Foundation Course for AIPMT/Olympiad : Class 9
- Science FR Ninth Class Part-3 (Biology) (PB)
- Physics Part 1 Class - 9
- Chemistry Part 2 Class - 9
- The IIT Foundation Series Physics Class 9
- The IIT Foundation Series Chemistry Class 9
- Oswaal CBSE CCE Question Banks With Complete Solution For Class 9 Term-II (October To March 2016) Science
- x am idea science class 9 term -2
- NCERT Exemplar Problems: Solutions Science Class 9
- NCERT Solutions - Science for Class IX

- ncert solutions for class 6 Science
- ncert solutions for class 6 Maths
- ncert solutions for class 7 Science
- ncert solutions for class 7 Maths
- ncert solutions for class 8 Science
- ncert solutions for class 8 Maths
- ncert solutions for class 9 Science
- ncert solutions for class 9 Maths
- ncert solutions for class 10 Science
- ncert solutions for class 10 Maths