# Rearranging equations.

I'm finnaly in physics and doing quite well but I am having some trouble.
d=Vi t + 1/2 a t^2 ,t=?
first I would divide both sides by Vi?
d/Vi=t + 1/2 a t^2
then divide both sides by 1/2 a? would it be d/Vi/a or d*a/Vi or d/Vi*a?
i'm going to just guess d/Vi*a (since m/s/s=m/s^2)
d/Vi*a=t +1/2 t^2 would t^2 still be multiplied by 1/2?
(square root of) d/Vi*a= t+1/2 t would the entire left side be square rooted or just the top?
t+1/2 t equals t i believe so t=(sq root of) d/Vi*a is the correct?

Second problem equation:
Vf^2=Vi^2 + 2 ad , d=?
first I would divide both sides by d?
Vf^2/2d=Vi^2 + 2a would it I still multiply d by 2?
here's were I get stuck, what do I do with Vf^2?

VietDao29
Homework Helper
$$d = v_it + \frac{1}{2}at^2$$ And you want to solve for t. Rearrange it as:
$$\frac{1}{2}at^2 + v_it - d = 0$$. And it's a quadratic equation.
$$ax^2 + bx + c = 0$$, a, b, c are already known, x is the unknown.
And its solution can be found by letting: $$\Delta = b ^ 2 - 4ac$$
If $$\Delta \geq 0$$, then the equation has solution(s):
$$x = \frac{-b \pm \sqrt{\Delta}}{2a}$$
If $$\Delta < 0$$, then the equation has no solution.
Here t is unknown.
---------------
$$v_f ^ 2 = v_i^2 + 2ad$$
Isolate d by adding $=v_i^2$ to both sides.
$$v_f ^ 2 - v_i^2 = 2ad$$
To isolate d, divide both sides by 2a:
$$\frac{v_f ^ 2 - v_i^2}{2a} = \frac{2ad}{2a}$$
$$\frac{v_f ^ 2 - v_i^2}{2a} = d$$
Viet Dao,

VietDao29 said:
$$d = v_it + \frac{1}{2}at^2$$ And you want to solve for t. Rearrange it as:
$$\frac{1}{2}at^2 + v_it - d = 0$$. And it's a quadratic equation.
$$ax^2 + bx + c = 0$$, a, b, c are already known, x is the unknown.
And its solution can be found by letting: $$\Delta = b ^ 2 - 4ac$$
If $$\Delta \geq 0$$, then the equation has solution(s):
$$x = \frac{-b \pm \sqrt{\Delta}}{2a}$$
If $$\Delta < 0$$, then the equation has no solution.
Here t is unknown.
---------------
$$v_f ^ 2 = v_i^2 + 2ad$$
Isolate d by adding $=v_i^2$ to both sides.
$$v_f ^ 2 - v_i^2 = 2ad$$
To isolate d, divide both sides by 2a:
$$\frac{v_f ^ 2 - v_i^2}{2a} = \frac{2ad}{2a}$$
$$\frac{v_f ^ 2 - v_i^2}{2a} = d$$
Viet Dao,

careful with the signs.

VietDao29
Homework Helper
Whoops, stupid typo... :grumpy: