H=54+1/2at^2 solve for a
Im lost on if I should square the 1/2at first?

Other answer formats are possible.

Explanation:

Let's say we want to solve the equation h = 54 + 10a for the variable 'a'. This is relevant to the problem at hand don't worry.

What we'd do is two things

So the steps look like this

h = 54 + 10a

h-54 = 10a

(h-54)/10 = a

a = (h-54)/10

Now let's say we wanted to solve this equation: h = 54 + ca, where c is a constant and can stand in place of any number. The steps would be pretty much identical. Instead of dividing both sides by 10, we divide both sides by c. We can do this as long as c isn't zero of course.

That means the equation h = 54+ca solves to a = (h-54)/c

Why did I do those examples? To build up to the equation your teacher gave you.

Let's replace c with 0.5t^2

This is because 1/2 = 0.5

The 1/2at^2 is the same as (1/2t^2)a

So instead of a = (h-54)/c, we'd get a = (h-54)/(0.5t^2)

As a last optional and encouraged step, we can multiply every term by 2 to get rid of that decimal value.

Doing so makes a = (h-54)/(0.5t^2) turn into a = (2h-108)/(t^2)

Here's another way to solve:

h = 54 + (1/2)at^2

h-54 = (1/2)at^2

2(h-54) = at^2

2h-108 = at^2

(2h-108)/(t^2) = a

a = (2h-108)/(t^2)

In the second step, I subtracted 54 from both sides. Immediately afterward, I multiplied both sides by 2 so I could clear out the fraction. Then I divided both sides by t^2 to fully isolated the variable 'a'.

b. hoped i'm right lol

step-by-step explanation: