I got an answer but I am not entirely sure if i was right. can someone please help to verify my answers. thanks
For a body falling under gravity in air, the rate of change of velocity is given by dV/dt = -k(V-P) where P and K are constants. This has a solution: V=P+Ae^(-kt). If the rate of change by dV/dt = -0.02(V-490), find:
a) an equation for V, remembering that initially V=0
b) the velocity after 10 seconds (2dp)
c) the maximum velocity which could be achieved, assuming no outside interference.
these are my answers
a) dV/dt = -k(V-P)
= -0.02(V-490)
Therefore: k=0.02 and P=490
V=P+Ae^(-kt)
k=0.02, P=490
V=490+Ae^(-0.02t)
when t=0, V=0
0=490+Ae^0
therefore A=490
V=490-490e^(-0.02t)
V=490[1-e^(-0.02t)]
b) when t=0,
V=490(1-e^-0.2)
V=88.82193099
c)maximum velocity when dV/dt=0
dV/dt=-0.02(V-490)
0=-0.02(V-490)
0=V-490
therefore V=490m/s
but from the equation:
V=490[1-e^(-0.02t)] the velocity cannot be equal to 490m/s as e^(any No.) cannot be equal to zero. So how is it that it can have a maximum velocity of 490m/s. What have i done wrong?
This is all correct.
This is correct too but remember that they are asking for the answer to 2 decimal places hence
It's an assumption. As , it will be a max. It will never approach 490 or exceed it as it is an asymptote so it is assumed to be a maximum (as it will continue 489.9999.....). See the graph attached, it would appear clearer visually.