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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.

Note by Soumo Mukherjee
3 years, 1 month ago

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