Our Problem can be classified as an indirect problem of kinematics.
Thus from **(1.3)** & **(1.2)** we have to move towards **(1.1)** .It is of utmost importance in any kind of problem to **know your direction** (even if you don't know the path.)

Using **(1.3)**

\[\cfrac { dv }{ dt } =a \\ dv=adt \\ \int _{ u }^{ v }{ dv } =\int _{ 0 }^{ t }{ adt }\]

As the time changes from **0** to **t** the velocity changes from **u** to **v**.So on the left hand side the summation is made over **v** from **u** to **v** whereas on the right hand side the summation is made on time from **0** to **t**. Evaluating the integrals we get

\[{ [v] }_{ u }^{ v }=a{ [t] }_{ 0 }^{ t } \\ v-u=at \\ v=u+at\]

Using **(1.2)** the last equation may be written as

\[\cfrac { dx }{ dt } =v=u+at \\ dx=(u+at)dt \\ \int _{ 0 }^{ x }{ dx } =\int _{ 0 }^{ t }{ (u+at)dt }\]

At **t=0** the particle is at **x=0**. As time changes from **0** to t the position changes from **0** to **x**.So on the left hand side the summation is made on position from **0** to **x** whereas on the right hand side the summation is made on time from **0** to **t**.Evaluating the integrals we get,

\[{ [x] }_{ 0 }^{ x }=\int _{ 0 }^{ t }{ udt } \int _{ 0 }^{ t }{ atdt } \\ x=u{ [t] }_{ 0 }^{ t }+a{ { [t }^{ 2 }/2] }_{ 0 }^{ t } \\ x=ut+(a{ t }^{ 2 })/2\]

Using the above two derived expressions,

\[{ v }^{ 2 }={ (u+at) }^{ 2 } \\ ={ u }^{ 2 }+2uat+{ a }^{ 2 }{ t }^{ 2 } \\ ={ u }^{ 2 }+2a[ut+\cfrac { 1 }{ 2 } a{ t }^{ 2 }] \\ ={ u }^{ 2 }+2ax\]

The equations

\[v=u+at\]

\[x=ut+\frac { 1 }{ 2 } a{ t }^{ 2 }\]

\[{ v }^{ 2 }={ u }^{ 2 }+2ax\]

are used very frequently while solving problems in kinematics involving constant acceleration as one of the physical quantities.If,however,the acceleration isn't constant,these three equations are not useful.We, then, need other tools & procedures.

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