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Brian M.
Tutor, mechanical engineer and road warrior
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Physics (Newtonian Mechanics)
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Question:

Elon Musk is testing a out a new reusable rocket. In order for the rocket to be reusable it needs to land ¨softly¨ (ie. the final velocity just above the surface needs to be 0). This is accomplished by igniting the thrusters while the rocket falls back to earth to slow the rocket down Given the initial velocity of the rocket is $$v_{i}$$ , and that the thrusters for landing produce a constant force of $$F_{thrust}$$ and can only be fired once, calculate the following. (assuming mass is constant) -What is the maximum height of the rocket -How far above the Earth should the thrusters fire ? -How long does the entire trip take ? Answer in terms of $$ m,g,F_{thrust}$$, and $$v_{i}$$ neglecting air resistance

Brian M.
Answer:

I love this question because it has so many different way to solve this , I will be using conservation of energy however it is just as possible using kinematic equations. I purposely left out values because the process is more important than plugging to get the final values Lets split the question into three parts and tackle each one in order. 1.)Max Height of the Rocket As mentioned previously the easiest/quickest way to solve this is to use energy balance. Lets do energy balance for right after take off and the highest point of the flight. From energy conservation: $$E_{init}=E_{apex}$$ Since the initial state only has kinetic energy and setting the earth as our reference point for potential energy we can see that the initial kinetic energy of the rocket changes forms to potential energy at the height of the flight. From our energy balance and some algebra we get the following expression $$KE_{init}=PE_{apex}$$ $$1/2v_{i}²=gh$$ (masses cancel out) $$\frac{v_{i}²}{2g}=h$$ which gives us our final result for part 1 2.) How far above the ground do we start the thrusters ? This part has a bunch of different way to be solved but the most efficient way is to employee energy conservation from the initial launch and the final landing. As the rocket flies skyward the kinetic energy changes to potential and vice versa on the way back down to earth. However when the rocket is safe and sound on the ground it has zero energy due to having no velocity and no height. So how was energy removed from the rocket ? The answer : Work So our energy balance looks like the following: $$E_{initial}-E_{final}=W_{net}$$ $$1/2mv_{i}²=W_{net}$$ Using the definition of work with constant force acting parallel to the direction of motion ($$W=F_{net}x$$) we get. $$1/2mv_{i}²=F_{net}x$$ $$F_{net}=F_{thrust}-mg$$ (from free body diagram Using algebra to solve for x (height above the earth to start firing) $$x=\frac{v_{i}²}{2(F_{thrust}/m-g)}$$ Which is how high above the ground the rockets should start firing for smooth landing 3.) Total time of flight -----Coming Soon------ I have the solution but just not the time currently to type it out :(

MATLAB
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Question:

Write a function that takes an integer as input and calculates the factorial using: 1.) a for loop 2.) a while loop 3.) recursion What are the associated worst case runtime of these functions ?

Brian M.
Answer:

Below is the example code , best viewed if copy and pasted into matlab . The runtime for all of the functions is $$O(n)$$ where n is the value of the input parameter, The all have to complete the same number of fundamental operations. I choose this example because it demonstrates some relatively simple but important fundamentals. -------Begin Code--------- %This is a nest function that is used to feed test cases, save us %function as a seperate function by itsself to be able to use from the %command line (not in debug mode) testN=4; %number for testing result = [factorial_for_loop(testN),factorial_while_loop(testN),factorial_recursion(testN)]; disp(result); function [while_loop_result] = factorial_while_loop(n) %using while loop for factorial %initial terms fact = n; while(n>1) fact = fact*(n-1); n=n-1; %decrement n end while_loop_result=fact; %return result end function [for_loop_result] = factorial_for_loop(n) %for loop method % terms natNums = 1:n; fact=1; for i=natNums %iterate through each natural number to n fact=fact*i; end for_loop_result=fact; %return result end function [recursion_result] = factorial_recursion(n) %recursion method if (n==1) %base case, recursion stops here recursion_result =1; else %continue deeper, inception style. n! = n*(n-1)! recursion_result =factorial_recursion(n-1)*n; end end

Calculus
TutorMe
Question:

Quaker has decided to make a parabola shaped bowl to improve math students cereal eating experience. Assuming the bowl as a circular cross section , develop a formula to determine the volume of the bowl. Assume the bowl´s spline is given by the equation $$y=bx²$$ Give the final answer in term of $$b$$ and the height of the bowl.

Brian M.
Answer:

To determine the volume we will have to use washer method. This entails ¨slicing¨ the shape into an infinite number of very very small cylinders using the power of calculus. First lets ¨sum¨ all the small volumes of our bowl to get the total volume $$V=\int{dV}$$ Next we want to construct our infinitesimally small cylinders or ¨washers¨ we will be adding together to get the total volume. Because we are designing a bowel that want to hold milk in the vertical direction , we will be ¨slicing¨ parallel to the x axis and stacking our infinitesimal cylinders in the y direction. Therefore our infinitesimal volume $$dV$$ can be expanded as $$dV = \pi r²dy$$ (infinitesimal cylinder) $$dV= \pi x²dy$$ (since the radius of cylinder is measured from y=0 (the y axis) the radius is just equivalent to x From our equation of the bowl we know $$y=bx²$$. Therfore by substituing in our value for $$x²$$ we get: $$dV=\pi \frac{y}{b} dy$$ Substitute into the integral $$V=\int{\pi \frac{y}{b} dy}$$ Add our bounds for $$y$$ $$V=\int_{0}^{y}{\pi \frac{y}{b} dy}$$ Then by evaluating the integral we achieve our final solution. Proving Quaker right once again that math can be fun. $$V=\frac{h^2}{2b}$$

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