+0

# Differential Equation

+1
1200
5
+105

Find the velocity of a falling body at t=5 seconds

Air resistance is proportional to the velocity of the body.

weight of the body=50kg

Height = 100m

Initial velocity = 5m/sec

Limiting velocity = 65/sec

Any help will be much appreciated :)

Feb 21, 2017

#3
+33617
+5

When you integrated to find t you put the constant inside the log  term. If you'd simply added the constant to the right hand side you would have found that c1 = 50ln(50g - 5k)/k and subsequently that k = 50g/65

.

Feb 21, 2017

#1
+33617
+5

"Find the velocity of a falling body at t=5 seconds

Air resistance is proportional to the velocity of the body.

weight of the body=50kg

Height = 100m

Initial velocity = 5m/sec

Limiting velocity = 65/sec"

I've assumed earth gravity above.  However, if you integrate again to find the height after 5 seconds you get a negative height, so perhaps this isn't on earth!  Also, air resistance is more normally taken to be proportional to velocity squared (though this means there is no nice closed form solution for velocity as a function of time - it has to be calculated numerically).

Feb 21, 2017
#2
+118623
0

Hi MarcoP,

I'll give it a shot, just don't rely on my answer :))

Find the velocity of a falling body at t=5 seconds

weight of the body=50kg

Air resistance is proportional to the velocity of the body.

weight of the body=50kg =m

\begin{align}\\air\;\; resistance\;\; up \;&\;\alpha \;\; velocity \;\;down\\ mass*accel\;&\;\alpha \;\; velocity \;down \\ m\frac{dv}{dt}&=k*\frac{dx}{dt} \\ \frac{dv}{dt}&=\frac{k}{m}*\frac{dx}{dt} \end{align}

Height = 100m

Initial velocity = 5m/sec

Limiting velocity = 65m/sec

Let downwards be positive velocity

$$Initially\\ \;\;t=0\;\;\;\;\ x=100 \;\; \;\; \dot x =5 \;\; \;\;\ddot x=g-\frac{k}{50}*5=g-0.1k$$

Ongoing

\begin{align}\\\frac{dv}{dt}&=g-\frac{k}{50}*v\\ \frac{dv}{dt}&=\frac{50g-kv}{50}\\ \frac{dt}{dv}&=\frac{50}{50g-kv}\\ t&=\frac{50ln(50g-kv+c_1)}{-k}\\~\\ &When\;\;t=0 \;\;v=5\;\;find\;\; c_1\\~\\ 0&=\frac{50ln(50g-5k+c_1)}{-k}\\ 0&=50ln(50g-5k+c_1)\\ 1&=50g-5k+c_1\\ c_1&=1-50g+5k\\ \frac{-kt}{50}&=ln(50g-kv+1-50g+5k)\\ \frac{-kt}{50}&=ln(1-kv+5k)\\ e^{\frac{-kt}{50}}&=1-kv+5k\\ &As \;\;t\rightarrow\infty \quad v\rightarrow 65\\~\\ 0&=1-65k+5k\\ 0&=1-60k\\ 60k&=1\\ k&=\frac{1}{60}\\ so\\ e^{\frac{-t}{60*50}}&=1-\frac{v}{60}+\frac{5}{60}\\ e^{\frac{-t}{3000}}&=\frac{13}{12}-\frac{v}{60}\\ &When\;\; t=5\\ e^{\frac{-5}{3000}}&=\frac{13}{12}-\frac{v}{60}\\ e^{\frac{-1}{600}}&=\frac{65}{60}-\frac{v}{60}\\ 60e^{\frac{-1}{600}}&=65-v\\ v&=65-60e^{\frac{-1}{600}}\\ v&=5.1 \end{align}\\

Which is wrong   :(

Alan can you see my error with any ease ?

*

Feb 21, 2017
#3
+33617
+5

When you integrated to find t you put the constant inside the log  term. If you'd simply added the constant to the right hand side you would have found that c1 = 50ln(50g - 5k)/k and subsequently that k = 50g/65

.

Alan  Feb 21, 2017
#4
+118623
0

Doh!

Thanks Alan, it is a log, I was multiplying not adding !!

It was so obvious as soon as you said it.

I'll try doing it again properly.  Justy got to get up to it.  :))

I hope you are having a really good birthday :))

Melody  Feb 21, 2017
#5
+105
0

Thanks Alan and Melody.

This will give me lots to chew on :)

Feb 21, 2017