Do I square acceleration when solving.

[mrserv0n]

Okay

X=Vo*t+1/2*a*t^2

x=2.80 m/s + 8.5s + .5 * 11.3 m/s^2 * 8.5s^2

Ok so I first do my exponent with time 8.5 * 8.5 = 72.25

but do I multiply that by 11.3 * .5 or do i have to square the 11.3 (acceleration) which is 127.69.

so is it 11.3 * .5 * 72.25 or 127.69 * .5 * 72.25?

 

[Stephan Hoyer]

You are simply plugging in the given values for V₀, t and a.

Your top equation, doing this replacement, looks like this:

x = (2.80 m/s) * (8.5s) + 1/2 * (11.3 m/s²) * (8.5 s)²

I don't know where you get the plus between the first two components, but that's multiplication.

Using the parentheses should make it clearer which values need to be squared and which do not.

A squared unit only refers to the unit itself being squared, not any values that use that unit. If the t² unit in acceleration always meant that the value for the acceleration should be squared, what sense would it make to state the original value?

 

[endeavor]

mrserv0n, don't start a new topic, look at your original thread.

you know 3 variables, x = 40.0m, t = 8.50s, and v = 2.80m/s
you need to find v₀, the initial speed, and a, the acceleration

use your knowledge of algebra to find these 2 variables in 2 equations