Calc1231W7T3 - Velocity of a falling object

Friday, 7 September 2018

5:24 PM

Machine generated alternative text: Suppose that v(t) is the velocity of a falling body depending on time t. The following differential equation dc ¯ g — kv, dt (where k and g are constants) can be solved using an integrating factor. To do this; rearrange equation (1) into the shape dc dt Then multiply both sides of this equation by the integrating factor kdt _ ekt Note that the constant of integration is typically omitted in this step. This results in an equivalent differential equation: dc kt kt dt Equation (3) can be rewritten in the form (you can check this; by using the product rule on the left hand sidey Integrating with respect to t we get kt ve for some constant C, and so the solution to equation (1) is g/k + Note: the question does not specify the initial velocity of this falling object. This will be reflected in your answer by the C term.

 

Machine generated alternative text: dv g — let' has a solution which was computed in the previous question as The equation dt g — kt This represents the velocity of a falling object, subject to gravity and air resistance As for the constants C, g, k: O depends on the gravity (which near the Earth is -9_8 m/sec), O depends on the air resistance, and c O depends on initial conditions. If the particle is initially at rest; meaning that when t = 0, v and so purely in terms of g and k, O , then we can deduce that

 

Machine generated alternative text: Suppose that the velocity v of a falling body, in some unit system; satisfies = 1—4t.'v and when t dv dt 4. Then we can solve that We can also solve for the terminal velocity. Notice that as t grows large, v approaches some limit; this limit is the terminal velocity. Calculate lim v 00 Challenge: Have a look at the general formula for t' that you calculated in the previous question; and determine if (or how) the terminal velocity depends on the constants g,k and C

 

Machine generated alternative text: We usually use Newton's laws and integration to determine position x as a function of time t. Suppose the falling body starts at a height h off the ground att by the differential equation dc = g — Jet' dt We have already seen how to derive 1 Now lets integrate once more, to find the actual position as a function of time: 0 and has velocity v given Now using the initial conditions, we find K So altogether - g/kA2 kt+e 'let 1)

 

Machine generated alternative text: In the 17th century it was widely believed that the velocity of a falling object was proportional to distance travelled_ Galileo discovered that this cannot be the case. Let's begin by supposing that velocity is proportional to the distance travelled from an initial resting point h meters from the ground. Then This differential equation is linear separable and has solution If the object is initially at height h, then when t dc k(h — x) where k 0_ dt h so

 

 

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