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.

2. Originally Posted by chrisgo

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.
so, would you mind posting what you did so that we can check it for you?

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?

4. Originally Posted by chrisgo

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)]
This is all correct.

Originally Posted by chrisgo
b) when t=0,
V=490(1-e^-0.2)
V=88.82193099
This is correct too but remember that they are asking for the answer to 2 decimal places hence $v=88.82ms^{-1}$

Originally Posted by chrisgo
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?
It's an assumption. As $e^t \to \infty$, 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.

5. Originally Posted by chrisgo

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 (take note of that..)

V=490-490e^(-0.02t) (correct!)
V=490[1-e^(-0.02t)]

b) when t=0, (oops, t=10)
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?
because the object is falling and as it continues to fall, the velocity continues to increase..

take note of that equation, as t tends to infinity, e^(-0.02t) goes to 0 and so V goes to 490.. this is what the blue statement says..