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Question: The motion of a body is given by the equation \(\frac{dv}{dt} = 6 - 3v\)where v is the speed in m s<...

The motion of a body is given by the equation dvdt=6−3v\frac{dv}{dt} = 6 - 3vwhere v is the speed in m s-1 and t is time in s. The body is at rest at t = 0. The speed varies with time as

A

v=(1−e−3t)v = (1 - e^{- 3t})

B

v=2(1−e−3t)v = 2(1 - e^{- 3t})

C

v=(1+e−2t)v = (1 + e^{- 2t})

D

v=2(1+e−2t)v = 2(1 + e^{- 2t})

Answer

v=2(1−e−3t)v = 2(1 - e^{- 3t})

Explanation

Solution

dvdt=6−3vordt=dv6−3v\frac{dv}{dt} = 6 - 3vordt = \frac{dv}{6 - 3v}

Integrating both sides, we get

t=−13in(6−3v)+Ct = - \frac{1}{3}in(6 - 3v) + C

Where C is a constant of integrations

At t = 0 , v = 0

∴C=13ln⁡6\therefore C = \frac{1}{3}\ln 6

∴t=−13ln⁡(6−3v6)\therefore t = - \frac{1}{3}\ln\left( \frac{6 - 3v}{6} \right)

−3t=ln⁡(6−3v6)- 3t = \ln\left( \frac{6 - 3v}{6} \right)

e−3t=1−12vorv=2(1−e−3t)e^{- 3t} = 1 - \frac{1}{2}vorv = 2\left( 1 - e^{- 3t} \right)