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THe motion of a body is given by the equation dv(t)/dt=6.0-3v(t), where v(t) is speed in m/s and t in sec. if a body was at rest at t=0

1. the terminal speed is 4m/s.
2. the speed varies with time as v(T)=2(1-e^(-5t)) m/s.
3. the speed is 0.1 m/s when the acceleration is half the initial value.
4. the magnitude of the initial acceleration is 6.0 m/s^2
4 years ago

by integration we get v=2-e-3t
so first 3 options or wrong
and  ans is 4 as intial a is 6
plz approve if useful and incase of any doubts ask in ans box
4 years ago

can u explain how did u get v=2-e^-3t
4 years ago

given dv(t)/dt=6.0-3v(t)
so dv(t)/6.0-3v(t)=dt
now take integral on both sides
so on ∫ we get v=2-e^-3t
PLZ APPROVE IF USEFULL

4 years ago

R U having problem in integration …...? or can u integrate the given function
if u dont no integration plz let me know i will help u out
4 years ago

i just know the basics of integration
i dont know how to integrate the above function
4 years ago

k i will help u on it  after 7;30 today
4 years ago

im really sorry but i have to do my class work for tmrw.
thanks for helping me
just write the explanation here after 7:30.
if i have any doubt in that i’ll post it here tmrw.
thank you so much
4 years ago

integration in lower levels is formula based so learning formulas is must
since u no basics u might know all formulas if u dont know them learn them
now u need to remember that
if ∫f(x)dx=g(x)+c then ∫f(ax+b)dx=(1/a)g(ax+b)+c _____________(1)
so now we no that ∫df(x)/f(x)=Inf(x)+c
here we have  dv/dt=6.0-3v
so  ∫dv/6.0-3v=∫dt  so from 1 we get
(-1/3)(log(6-3v)-log(6)=t
(u need to  take limits from o to v on LHS  and 0 to t on RHS)
now log (1-(v/2))=-3t
so e-3t =1-(v/2)
so v=2(1-e-3t )
PLZ APPROVE if usefull in  case of any doubs ask  in ans box

4 years ago

thank u so much for taking the trouble.
i totally understood it.

4 years ago

ur welcome
its my plessure
4 years ago
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