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An x-t graph shows displacement over time, and a v-t graph shows velocity over time. The combination of the two graphs can give you great detail about the motion of an object over a given period of time. For example, if an object moved 2 cm over 2 seconds on the x-t graph, that says nothing about what direction the object moved in, but if you combine that data with the v-t graph and see that over those 2 seconds the object had a positive acceleration, that means that the object was moving away from the origin of the graph.
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Help An object is thrown straight down from the top of a 300ft building with an initial velocity of -20 feet per second What is its velocity after 2 seconds and what it velocity after falling 150ft?
Given that dv/dt = a, you can take the integral of a, and get...vt = a t + C. Substitute v0 for C, and you get vt = a t + v0Assuming that the acceleration due to gravity is about -32 fps2, then the velocity after 2 seconds of an object down with an initial velocity of -20 fps is -80 fps.To solve the second part of the question, "what is the velocity after falling 150 f?", you need to integrate again, and get...xt = 1/2 a t2 + v0 t + C. Substitute x0 for C, and you get xt = 1/2 a t2 + v0 t + x0Solve for t, with xt = 150, (300-150), v0 = -20, and x0 = 300, using the quadratic equation, and you get t = 2.28 s. (You also get t = -1.03 s, but you can ignore that because it is the solution for time prior to the release point.)Go back to the first integral and plug in t = 2.28 s, and you get v = 92.9 fps.This answer assumes no resistance from velocity in air. At 92.9 fps, you might get some effect, depending on the shape and density of the object, so the real answer is less than stated.
Moment generating and the cumulant generating function of poisson distribution?
The moment generating function is M(t) = Expected value of e^(xt) = SUM[e^(xt)f(x)] and for the Poisson distribution with mean a inf = SUM[e^(xt).a^x.e^(-a)/x!] x=0 inf = e^(-a).SUM[(ae^t)^x/x!] x=0 = e^(-a).e^(ae^t) = e^[a(e^t -1)]
How to find the impulse of 20 units?
t=0:.02:20 > num1=[1] > num2=[1 0] > denum=[.2 .2 6] > sys1=tf(num1,denum) > sys2=tf(num2, denum) > xt=impulse(sys1,t) > xdott=impulse(sys2,t) > plot(t,xt,'r',t,xdott,'b') >
How do you find the endpoint of a line segment when one endpoint is given and a point one third of the way is given?
There are two possible answers given the information. What isn't given is if the second point is one third of the way from the known or unknown endpoint. Say the known endpoint is (xe,ye) and the point one third of the way along is (xt,yt). If the point one third of the way is closest to the known endpoint, the other endpoint would be (xe+3*(xt-xe), ye+3*(yt-ye)). This is probably the answer implied by your question. If the point is closest to the unknown endpoint the the unknown endpoint is (xe+(3/2)*(xt-xe), ye+(3/2)*(yt-ye)).
How do you show a variance matrix is nonnegative definite?
I assume you mean covariance matrix and I assume that you are familiar with the definition:C = E[(X-u)(X-u)T]where X is a random vector and u = E(X) is the meanThe definition of non-negative definite is:xTCx ≥ 0 for any vector x Є RSo is xTE[(X-u)(X-u)T]x ≥ 0?Then, from one of the covariance properties:E[(xT(X-u))((X-u)Tx)] = E[xT([(X-u)(X-u)T]x)] = E[((xTI)x)] = E[xTx]Finally, since we've already defined x to have only real values, xTx is therefore non-negative definite by definition.
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