Question

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

Answer 1

The minimum velocity of the particle =
`-e^(-2 )` units

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) =
`(d)/(dt)(40 + te^{-(t)/(20) } )`

Now,

`(d)/(dt)(40 + te^{-(t)/(20) } )` =
`(d)/(dt)(40 ) + (d)/(dt)(te^{-(t)/(20) } )`

= 0 +
`t(d)/(dt)(e^{-(t)/(20) } ) + e^{-(t)/(20) }(d)/(dt)(t )`

=
`t(-(1)/(20) )e^{-(t)/(20) } + e^{-(t)/(20) }`

⇒v(t) =
`-(t)/(20)e^{-(t)/(20) } + e^{-(t)/(20) }`

Now,

For minimum velocity, Put
`(d)/(dt)(v(t)) = 0`

Now,

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

=
`-(2)/(20) e^{-(t)/(20) } + (t)/(400) e^{-(t)/(20) }`

Now,

Put
`(d)/(dt)(v(t)) = 0`, we get

`-(2)/(20) = - (t)/(400)`

⇒t = 40

Now,

Check that the point is minimum or maximum

Calculate
`(d^(2) )/(dt^(2) ) [v(t)]`

Now,

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

=
`(1)/(400)e^{- (t)/(20) } [ 3 - (t)/(20)]`

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

∴ we get

t = 40 is point of minimum

So,

The minimum velocity be

v(40) =
`-(40)/(20)e^{-(40)/(20) } + e^{-(40)/(20) }`

=
`-2e^(-2 ) + e^(-2 )`

=
`-e^(-2 )`

⇒v(40) =
`-e^(-2 )` units

∴ we get

The minimum velocity of the particle =
`-e^(-2 )` units