Question

Solve the following two equations for the (positive) time, t, and the position, x. Assume SI units.

Answer 1

A system of equations is a group of equations with the same variables that we need to solve simultaneously. Such that the solutions is given by the intersection betweens graphs of the functions.

We will see that the solution is (16.4, 807.0)

Here the system is:

$x = 3.00*t^2\\x = 45.0*t + 69.0$

To solve a system we usually need to isolate one of the variables in one equation and replace that in the other equation, here we already see that we have x isolated in the two equations, so we can write:

$3.00*t^2 = x = 45.0*t + 69.0\\3.00*t^2 = 45.0*t + 69.0$

Now we can solve the above equation for t:

$3.00*t^2 = 45.0*t + 69.0\\\\3.00*t^2 - 45.0*t - 69.0 = 0$

This is just a quadratic equation, the solution is given by the Bhaskara's formula, we will get:

$t = \frac{-(-45.0) \pm \sqrt{(-45.0)^2 - 4*(3.00)*(-69.0)} }{2*3.00} \\\\t = \frac{{(45.0) \pm 53.4} }{6.00}$

Then the two values of t are:

t = (45.0 + 53.4)/6 = 16.4

t = (45.0 - 53.4)/6 = -1.4

We want the positive solution, so we choose t = 16.4

To complete the solution we need to evaluate one of our functions in this time. Let's use the first one:

x = 45.0*16.4 + 69.0 = 807.0

Then the solution is:

(16.4, 807.0)

If you want to learn more you can read:

brainly.com/question/12895249