Tutor profile: Mitchell S.

A stunt man drives a car off of a 10.0 m high cliff at a speed of 20.0 /s. How far does the car land from the base of the cliff?

First, draw a simple picture to represent this scenario. Assign the x-y axis to where the car is at the beginning of the problem, right before it drives off of the cliff.. Now, we write out our givens: t_0 = 0 s x_0 = 0 m y_0 = 0 m v_0y = 0 m/s v_0x = 20 m/s a_x = 0 m/s2 a_y = -9.8 m/s2 y_1 = -10 m We now write down what we are being asked to find: x_1 = ? We can recognize that once the car goes off of the cliff it will obey the rules set forth for projectile motion. We can solve this sort of problem using the kinematic equations. We know that the initial x position is zero, and that acceleration in the y-direction is solely due to gravity, so our equations reduce to: x_1 = x_0 + v_0x*t = v_0x*t y_1 = -10 m = y_0 + v_0y*t + 0.5*g*t^2 => y_1 = - 0.5*g*t^2 Now we can use the second equation to solve for time t = sqrt(-2y_1/g) = sqrt(-2*-10/9.81) = 1.43 s This can then be plugged into the first equation to provide us with the horizontal distance travelled: x_1 = v_0x*t = 20*1.43 = 28.6 m

You are asked to find the roots of a function: f(x) = exp(-x) - x Find the roots using the Newton- Raphson method.

function [root,ea,iter]=newtraph(func,dfunc,xr,es,maxit,varargin) % newtraph: Newton-Raphson root location zeroes % [root,ea,iter]=newtraph(func,dfunc,xr,es,maxit,p1,p2,...): % uses Newton-Raphson method to find the root of func % input: % func = name of function % dfunc = name of derivative of function % xr = initial guess % es = desired relative error (default = 0.0001%) % maxit = maximum allowable iterations (default = 50) % p1,p2,... = additional parameters used by function % output: % root = real root % ea = approximate relative error (%) % iter = number of iterations if nargin<3,error('at least 3 input arguments required'),end if nargin<4|isempty(es),es=0.0001;end if nargin<5|isempty(maxit),maxit=50;end iter = 0; while (1) xrold = xr; xr = xr - func(xr)/dfunc(xr); iter = iter + 1; if xr ~= 0, ea = abs((xr - xrold)/xr) * 100; end if ea <= es | iter >= maxit, break, end end root = xr;

A rectangular field must be enclosed with a fence. There is 400 feet of fencing material and a building is on one side of the field (It doesn't require any fencing). Determine the dimensions of the field that will enclose the largest area

This is an optimization type of problem. Start by determining what the goal is (maximize the area) and what the constraint is (the amount of fencing). Maximize: A =L*W Constraint: 400 = 2*L + W (Because there is one side with a building) Solve the constraint equation for width (W): W = 400 - 2L Substitute the constraint into the area function: A(L) = L * (400 - 2L) = 400L - 2L^2 Now the area (A) is a function of only length. This can be differentiated and the critical points determined. The critical points will help us find the maximum for this problem A'(L) = 400 - 4L Setting this value equal to zero helps us find a critical point of L = 100. We can then plug this back into our original constraints equation to determine the width that provides us with the maximum area. 400 = 2*100 + W This yields a width (W) of 200.