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taurus [48]
3 years ago
13

A particle moves along a horizontal line so that its position at time t, t ≥ 0, is given by

Mathematics
1 answer:
EastWind [94]3 years ago
4 0

Answer:

The minimum velocity of the particle  = -e^{-2 } units

Step-by-step explanation:

Given - A particle moves along a horizontal line so that its position at time t,

t ≥ 0, is given by  s(t) = 40 + te^−t/20.

To find - Find the minimum velocity of the particle for 0 ≤ t ≤ 100.

Proof -

Velocity, v(t)  = \frac{d}{dt}(40 + te^{-\frac{t}{20} } )

Now,

\frac{d}{dt}(40 + te^{-\frac{t}{20} } ) =  \frac{d}{dt}(40 ) + \frac{d}{dt}(te^{-\frac{t}{20} } )

                       = 0 + t\frac{d}{dt}(e^{-\frac{t}{20} } ) + e^{-\frac{t}{20} }\frac{d}{dt}(t )

                       = t(-\frac{1}{20} )e^{-\frac{t}{20} } +  e^{-\frac{t}{20} }

⇒v(t) = -\frac{t}{20}e^{-\frac{t}{20} } + e^{-\frac{t}{20} }

Now,

For minimum velocity, Put \frac{d}{dt}(v(t)) = 0

Now,

\frac{d}{dt}[v(t)]  = \frac{d}{dt} [ -\frac{t}{20}e^{-\frac{t}{20} } + e^{-\frac{t}{20} } ]

           = -\frac{2}{20} e^{-\frac{t}{20} } + \frac{t}{400} e^{-\frac{t}{20} }

Now,

Put \frac{d}{dt}(v(t)) = 0, we get

-\frac{2}{20} = - \frac{t}{400}

⇒t = 40

Now,

Check that the point is minimum or maximum

Calculate  \frac{d^{2} }{dt^{2} } [v(t)]

Now,

\frac{d^{2} }{dt^{2} } [v(t)] = \frac{d}{dt} [ -\frac{2}{20} e^{-\frac{t}{20} } + \frac{t}{400} e^{-\frac{t}{20} }]

            = \frac{1}{400}e^{- \frac{t}{20} } [ 3 - \frac{t}{20}]

⇒\frac{d^{2} }{dt^{2} } [v(t)]  = \frac{1}{400}e^{- \frac{t}{20} } [ 3 - \frac{t}{20}]  > 0

∴ we get

t = 40 is point of minimum

So,

The minimum velocity be

v(40) = -\frac{40}{20}e^{-\frac{40}{20} } + e^{-\frac{40}{20} }

      = -2e^{-2 } + e^{-2 }

      = -e^{-2 }

⇒v(40) = -e^{-2 } units

∴ we get

The minimum velocity of the particle  = -e^{-2 } units

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