I see that this is a quadratic equaion and it has been a while sense I have taken algebra 1.... I don\'t know how to solve for t3...
-2^-1 t3^2 - 180 s t3 + t3^2 = a^-1 (Vo (180 s) - 1,100 m + Xo)
how do I solve this for t3
the 3 is a subscript to t
180 s is 180 seconds and is considered one term
Vo is one term
Xo is one term
- 1,100 m is one term and is - 1, 100 meters
by the way i realized i could just graph the two equations...
were
a = .20
Vo = 5.494
but how do I write it for t3
To solve the quadratic equation for t3, you can follow these steps:
Step 1: Simplify the equation
Rearrange the equation to bring all terms to one side. In this case, move all terms to the left side of the equation. Additionally, convert any negative exponents into positive exponents:
-2^(-1)t3^2 -180st3 + t3^2 - a^(-1)(Vo(180s) - 1,100m + Xo) = 0
Step 2: Combine like terms
Combine the terms with similar variables and exponents:
(-2^(-1) + 1)t3^2 - 180st3 - a^(-1)(Vo(180s) - 1,100m + Xo) = 0
Simplifying further:
(-1/2)t3^2 - 180st3 - a^(-1)(Vo(180s) - 1,100m + Xo) = 0
Step 3: Set up the quadratic equation
The equation is now in the form Ax^2 + Bx + C = 0, where A = -1/2, B = -180s, and C = -a^(-1)(Vo(180s) - 1,100m + Xo).
Step 4: Solve the quadratic equation
To solve the quadratic equation, you can use various methods, such as factoring, completing the square, or using the quadratic formula. Here, we will use the quadratic formula:
The quadratic formula states that for an equation of the form Ax^2 + Bx + C = 0, the solutions for x are given by:
x = (-B ± √(B^2 - 4AC)) / (2A)
In our case, A = -1/2, B = -180s, and C = -a^(-1)(Vo(180s) - 1,100m + Xo). Therefore, the solutions for t3 are:
t3 = [180s ± √((180s)^2 - 4(-1/2)(-a^(-1)(Vo(180s) - 1,100m + Xo)))] / (-2 * (-1/2))
Simplifying further and rearranging:
t3 = [180s ± √(32a^(-1)(Vo(180s) - 1,100m + Xo) + (180s)^2)] / (-1)
So, by substituting the values of a, Vo, s, and Xo, you can calculate the values of t3 using the above formula.