resisted motion question (1 Viewer)

[a-n-d-r-e-w]

im having quite a bit of trouble with this question for some reason. Its in the new senior mathematics fitzpatrick boof (ex 35b, 4), and i dont understand how they get their answer.

'A particle of mass 10kg falls from rest and is subject to a force of (98 - 2v) newton where v is the speed at time t seconds. Find:
i) an expression for its momentum at time t
ii) its terminal velocity, i.e. its velocity as t -->infinity
iii) the distance fallen in the first 5 seconds'

the answers in the back are:
i) 490(1 - e^-0.2t)
ii) 49m/s
iii) 90m

 

[vds700]

im having quite a bit of trouble with this question for some reason. Its in the new senior mathematics fitzpatrick boof (ex 35b, 4), and i dont understand how they get their answer.

'A particle of mass 10kg falls from rest and is subject to a force of (98 - 2v) newton where v is the speed at time t seconds. Find:
i) an expression for its momentum at time t
ii) its terminal velocity, i.e. its velocity as t -->infinity
iii) the distance fallen in the first 5 seconds'

the answers in the back are:
i) 490(1 - e^-0.2t)
ii) 49m/s
iii) 90m

i)R = 10a = 98 - 2v
a = 9.8 - 0.2v
dv/dt = 9.8 - 0.2v
dt/dv = 1/(9.8 - 0.2v)
t = -5ln(9.8 - 0.2v) + c
when t = 0, v = 0, c = 5ln(9.8)
t = 5ln[9.8/(9.8 - 0.2v)]
0.2t = ln[9.8/(9.8 - 0.2v)]
e^0.2t = 9.8/(9.8 - 0.2v)
e^-0.2t = 9.8 - 0.2v / 9.8
9.8e^-0.2t = 9.8 - 0.2v
0.2 v = 9.8(1 - e^-0.2t)
v = 49(1 - e^-0.2t)
p = mv
= 490(1 - e^-0.2t)

ii)as t -> infinity, e^-0.2t -> 0, so v = 49 (1) = 49 m/s

iii)just integrate v then sub in 5 and u should get the answer

 

[a-n-d-r-e-w]

thanks heaps man, it seems really obvious now.