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)

dvdt=adv=adtuvdv=0tadt\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]uv=a[t]0tvu=atv=u+at{ [v] }_{ u }^{ v }=a{ [t] }_{ 0 }^{ t } \\ v-u=at \\ v=u+at

Using (1.2) the last equation may be written as

dxdt=v=u+atdx=(u+at)dt0xdx=0t(u+at)dt\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]0x=0tudt0tatdtx=u[t]0t+a[t2/2]0tx=ut+(at2)/2{ [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,

v2=(u+at)2=u2+2uat+a2t2=u2+2a[ut+12at2]=u2+2ax{ 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+atv=u+at

x=ut+12at2x=ut+\frac { 1 }{ 2 } a{ t }^{ 2 }

v2=u2+2ax{ 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
4 years, 11 months ago

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