# Tutor profile: Zack H.

## Questions

### Subject: Basic Math

For the following expression, solve in steps to show proper PEMDAS application. 12 - 1 * 0 + 4 / 2

1. 1 * 0 = 0; 12 - 0 + 4 / 2 2. 4 / 2 = 2; 12 - 0 + 2 3. 12 - 0 + 2 = 14

### Subject: Physics

A baseball player is running toward home plate to score the winning run. He begins his slide to the plate at 25 ft/sec, 8ft from the plate. He decelerates at a rate of 40 ft/sec^2. Assume that: - If the runner reaches or passes the plate, he will be making contact with the plate (and scores the run) - The runner is tagged (does not score the run) by the catcher just after he stops sliding Does he score the run and win the game? If not, how far short is the runner when he is tagged?

We know the following information from the problem statement: $$ $$ \begin{align*} Initial Speed\\ V_{0} &= 25 ft/sec\\ Acceleration\\ a &= -40 ft/sec^2\\ \end{align*} Let's also define the locations of the runner as $$x_{0} = 0 ft$$ and $$x = 8 ft$$. We can use the following equation of motion to determine the runner's sliding velocity at home plate: \begin{align*} V^2 = V_{0}^2 + 2a(x-x_{0}) \end{align*} Let's use our defined values and see what happens with the runner. Inserting the values above, we have: \begin{align*} V^2 &= (25 ft/sec)^2 + 2 (-40 ft/sec^2) (8 ft - 0 ft)\\ V^2 &= 625 ft^2/sec^2 + (-640 ft^2/sec^2)\\ V^2 &= -15 ft^2/sec^2\\ V &= \sqrt{-15 ft^2/sec^2}\\ \end{align*} We can see that the solution for the velocity at the end of the runner's 8 ft slide would be an imaginary number. This means that the runner does not slide the full 8 ft to home plate. Unfortunately, the runner does not score the run and win the game. To see just how close the runner was to scoring the run, we can go back to our equation of motion and redefine some of the values. In this case, we will redefine the final velocity, $$V$$, to be: $$V = 0 ft/sec$$ to represent the end of his slide. We will also change $$x$$ back to an unknown variable, because we know he does not slide the full 8 ft. Our equation becomes: \begin{align*} V^2 &= V_{0}^2 + 2a(x-x_{0})\\ (0 ft/sec)^2 &= (25 ft/sec)^2 + 2 (-40 ft/sec^2) (x - 0 ft)\\ \end{align*} Rearranging and solving for $$x$$: \begin{align*} 0ft^2/sec^2 &= 625 ft^2/sec^2 - 80x ft/sec^2\\ 80x ft/sec^2 &= 625 ft^2/sec^2\\ x &= 7.8 ft\\ \end{align*} This means the runner's slide extended 7.8 ft towards home plate. Thus, the runner was just 0.2 ft shy of scoring the game winning run!

### Subject: Algebra

Your friends, Tim and Katie, are on a game show. For a chance to win $20,000, they are each given one equation and asked to solve for 'x' and 'y'. They have a "phone a friend" lifeline that they can use. Luckily for them, their friend (you) is an algebra whiz and can help them win without breaking a sweat. The equations are as follows: \begin{align*} 8x + 4y &= 32\\ 4x - 2y &= 8 \end{align*} Solve for 'x' and 'y' so Tim and Katie can go on a lavish vacation. $$ $$

To begin, let's simplify the first equation in the system. Rearranging for y: \begin{align*} 8x + 4y &= 32\\ 4y &= 32 - 8x\\ y &= 8 - 2x\\ \end{align*} Next, let's use this definition for y in the second equation: \begin{align*} 4x - 2y &= 8\\ \end{align*} Replacing y, \begin{align*} 4x - 2 ( 8 - 2x) &= 8\\ \end{align*} Now we just simplify and solve for x! \begin{align*} 4x - 16 + 4x &= 8\\ 8x &= 24\\ x &= 3\\ \end{align*} So, now we know that $$x=3$$ and we can use this to solve for y in the simplified first equation, like so: \begin{align*} y &= 8 - 2x\\ y &= 8 - 2(3)\\ y &= 8 - 6\\ y &= 2\\ \end{align*}