Dynamics engineering question (1 Viewer)

[Bdogz]

can any1 help me with this question plz: a small projectile is fired vertically downward into a fluid medium with an intial velocity of 60m/s. Due to the resistance of the fluid the projectile experiences a deceleration equal to a=(-0.4v^3)m/s^2, where v is in m/s. Determine the projectile's velocity and postion 4 seconds after it is fired.

the answers are velocity=0.559m/s downwards, postion=4.43m (i dont knoe how 2 get to the answer lol)

 

[tommykins]

Show us what you have, then we'll help you.

Start off with
F = ma -mkv^3

 

[Bdogz]

well i sed a=dv/dt=-0.4v^3 and then i tried to take the integral of both sides so i can get velocity=something but im not sure what to take the integral with respect to, n if u say start off with F=ma-mkv^3 how do u knoe wat the mass is?

 

[sam0206]

To find velocity:
a=dv/dt=-0.4v^3
integral of dv/-0.4v^3
1/-0.4(1/-2)1/v^2 with integral limit things being 0 and 60= t-0
1/0.8(1/v^2-1/60^2)= t
v= ((1/60^2+0.8t)^-1/2)m/s
when t=4s,
v=0.559m/s downwards

To find position:
v=sd/dt= ((1/60^2+0.8t)^-1/2)
integral of (1/60^2+0.8t)^-1/2dt
s=2/0.8(1/60^2+0.8t)^1/2 limits 0 and t
s=1/0.4((1/60^2+0.8t)^1/2-1/60)m
when t= 4s,
s=4.43m

Dont know how it works but hopefully thats easy enough 4 u 2 read Bdogz i dont understand where the (1/-2) comes from wen working out the velocity. confusing

 

[tommykins]

you don't need mass in the equation , but it's a way to start off? I'm assuming theres an acceleration otherwise F = -mkv^3 would be the only thing you need. the k = 0.4 but putting a k there reduces decimals.

F = ma -> a = F/m -> sub in F and you get an equation for acceleration

dv/dt = -kv^3
dv/v^3 = -k dt
-1/2v^3 = -kt + c t = 0 , v= 60

so on so forth

 

[Bdogz]

you don't need mass in the equation , but it's a way to start off? I'm assuming theres an acceleration otherwise F = -mkv^3 would be the only thing you need. the k = 0.4 but putting a k there reduces decimals.

F = ma -> a = F/m -> sub in F and you get an equation for acceleration

dv/dt = -kv^3
dv/v^3 = -k dt
-1/2v^3 = -kt + c t = 0 , v= 60

so on so forth

k thnx dude that clears up a lot of things, appreciate the help

 

[Bdogz]

To find velocity:
a=dv/dt=-0.4v^3
integral of dv/-0.4v^3
1/-0.4(1/-2)1/v^2 with integral limit things being 0 and 60= t-0
1/0.8(1/v^2-1/60^2)= t
v= ((1/60^2+0.8t)^-1/2)m/s
when t=4s,
v=0.559m/s downwards

To find position:
v=sd/dt= ((1/60^2+0.8t)^-1/2)
integral of (1/60^2+0.8t)^-1/2dt
s=2/0.8(1/60^2+0.8t)^1/2 limits 0 and t
s=1/0.4((1/60^2+0.8t)^1/2-1/60)m
when t= 4s,
s=4.43m

Dont know how it works but hopefully thats easy enough 4 u 2 read Bdogz i dont understand where the (1/-2) comes from wen working out the velocity. confusing

wow detailed answer, thnx a lot u sound smart sam0206

 

[sam0206]

wow detailed answer, thnx a lot u sound smart sam0206

anytime

 

[tommykins]

k thnx dude that clears up a lot of things, appreciate the help

reason i didn't give you a solution was because it's better off if you did it yourself after being nudged in the right direction.

good luck! next time if you have a q, post what you have so far - people are more inclined to help if you've made some effort solving it.