Question

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

Answer 1

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