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)
Therefore: k=0.02 and P=490
when t=0, V=0
b) when t=0,
c)maximum velocity when dV/dt=0
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?