How do you solve for a: in #y= (at²)÷2 + vt#?

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Answer 1 · 2018-03-23T17:24:40

#a=(2(y-vt))/t^2#

The function #y=(at^2)-:2+vt# represents distance covered by an object starting with an initial velocity #v# with an accelaration #a#, in time #t#.

As #y=(at^2)-:2+vt#

#(at^2)-:2=y-vt#

i.e. #at^2=2(y-vt)#

and #a=(2(y-vt))/t^2#

Answer 2 · 2018-03-23T17:57:22

#a=(−2tv+2y)/t^2#

Step 1: Simplify

# y=(at^2)/2+vt#
Step 2: Flip the equation.

#1/2 at^2+tv=y#
Step 3: Add -tv to both sides.

#1/2 at^2+tv+color(red)(−tv)=y+color(red)(−tv)#
#1/2at2=−tv+y#

Step 4: Divide both sides by #t^2/2#.
#(1/2at^2)/color(red)(t^2/2)=(−tv+y)/color(red)(t^2/2)#
#a=(−2tv+2y)/(t^2)#