How do you simplify #\frac{at^{2}}{2}+ut#?

1 Answer
Nov 1, 2017

Answer:

#(at^2)/2+ut=barvxxt#, where #barv# is average velocity.

Explanation:

We can write #(at^2)/2+ut# as #(at^2+2ut)/2#, if you mean so the question has been answered.

This represents the distance #S# covered an object with an initial velocity #u# and a uniform accelaration #a# in time #t# i.e. #S=(at^2)/2+ut# or #S=(at^2+2ut)/2#.

It may be noted that if initial velocity is #u# and uniform accelaration is #a#, then the velocity after #t# units of time #v# is given by

#v=u+at# and average velocity #barv# is given by #barv=(v+u)/2# i.e.

but as #S=(at^2+2ut)/2=(t(at+u+u))/2=(t(v+u))/2#

or we can say #S=barvxxt#, where #barv# is average velocity.

Note that the different formulas are used in different circumstances and though #(at^2)/2+ut# simplifies to #barvxxt#, these are used depending on what variables are given to us, regarding motion with uniform accelaration. One more important formula is #v^2-u^2=2aS#.