Question #743f5

Question

1238 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-05T15:06:15

Because they are derived,on the basis of the consideration that acceleration is constant.

Suppose, acceleration = #a# =constant

So, #(dv)/(dt) = a# (as,acceleration = rate of change in velocity)

Or, #dv = adt#

Integrating from #v= u to v# and #t=0 to t#,
we get,

#v-u = at# or, #v=u+at#

Again,acceleration can be written as, #v(dv)/(dx) =a#

or, #v dv = adx#

Integrating from, #v=u to v# and #s=0 to s# we get,

#v^2-u^2 = 2as#

Or, #v^2=u^2 +2as#

Again, #(dx)/(dt) = u+at# (as, velocity = rate of change in displacement)

So, #dx = udt +atdt#

Integrating from, #x=0 to x# and, #t=0 to t#,we get,

#x = ut +1/2at^2#