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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