How do you solve for #a# in #y=v_0t+1/2at^2#?

1 Answer
Morgan
Dec 30, 2016

#a=(2(y-v_0t))/t^2#. See below.

Explanation:

We can manipulate this kinematic equation to solve for acceleration with a little algebra.

On the right side, we have the sum of #v_0t+1/2at^2#. We want to isolate #a#. First, we subtract #v_0t# from both sides.

#y-v_0t=cancel(v_0t)+1/2at^2cancel(-v_0t)#

#=>y-v_0t=1/2at^2#

We subtracted #v_0t# because we were adding it on the right side. We're left with only multiplication on the right. We can divide both sides by #t^2#.

#(y-v_0t)/t^2=(1/2acancel(t^2))/cancel(t^2)#

#=>(y-v_0t)/t^2=1/2a#

Now we divide both sides by #1/2#, or equivalently, multiply both sides by #2#.

#(2(y-v_0t))/t^2=1/cancel2a*cancel2#

#=>(2(y-v_0t))/t^2=a#

We can now calculate acceleration:

#a=(2(y-v_0t))/t^2#

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