A car is travelling at 100 km/h on a level road when it runs out of fuel. Its speed v (in km/h) star

Kyle Sutton 2022-07-14 Answered
A car is travelling at 100 km/h on a level road when it runs out of fuel. Its speed v (in km/h) starts to decrease according to the formula
d v d t = k v ( 1 )
where k is constant. One kilometre after running out of fuel its speed has fallen to 50 km/h. Use the chain rule substitution
d v d t = d v d s d s d t = d v d s v
to solve the differential equation.
Note: Although I haven't solved it yet, the answers say that this isn't a reasonable model as the velocity is always positive; I didn't make a typo in the question.
What I'm trying to do is solve velocity as a function of displacement (s, in km), velocity as a function of time (t, in hours), and displacement as a function of time (I need these functions for later parts of the question).
So far I've found velocity as a function of displacement (v(s)):
d v d t = k d s d t (from (1))
d v d t d t = k d s d t d t
v ( s ) = k s + C
v ( 0 ) = 100 C = 100 ,   v ( 1 ) = 50 k = 50
v ( s ) = 50 s + 100
Then I've tried to find velocity as a function of time (v(t)), but I've got stuck. I can't find any differential equation I can use to get this, or to get displacement as a function of time (s(t)).
The answer key says v ( t ) = 100 e 50 t and s ( t ) = 2 ( 1 e 50 t )
I've solved such questions many times before, but it's been a while so I'm a bit rusty. So, even a hint might be enough for me to realise what to do.
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Answers (1)

poquetahr
Answered 2022-07-15 Author has 18 answers
We can find v(t), taht k=50.
d v d t = k t = 50 t
v ( t ) = A e 50 t ,   for some constant  A
v ( 0 ) = 100 A = 100
v ( t ) = 100 e 50 t
We can find s(t) using d s d t and the chain rule, however I'm not sure if this is the most efficient method. We know from v(s) that d v d s = 50, so d s d v = 1 50 . We can also see from v(t) that d v d t = 5000 e 50 t .
d s d t = d s d v d v d t = 1 50 × 5000 e 50 t = 100 e 50 t
d s d t d t = 100 e 50 t d t
s ( t ) = 100 × 1 50 e 50 t + C
s ( t ) = 2 e 50 t + C
s ( 0 ) = 0 C = 2
s ( t ) = 2 e 50 t + 2 = 2 ( 1 e 50 t )
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