Respuesta :
Answer:
The minimum velocity of the particle = [tex]-e^{-2 }[/tex] 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) = [tex]\frac{d}{dt}(40 + te^{-\frac{t}{20} } )[/tex]
Now,
[tex]\frac{d}{dt}(40 + te^{-\frac{t}{20} } )[/tex] = [tex]\frac{d}{dt}(40 ) + \frac{d}{dt}(te^{-\frac{t}{20} } )[/tex]
= 0 + [tex]t\frac{d}{dt}(e^{-\frac{t}{20} } ) + e^{-\frac{t}{20} }\frac{d}{dt}(t )[/tex]
= [tex]t(-\frac{1}{20} )e^{-\frac{t}{20} } + e^{-\frac{t}{20} }[/tex]
⇒v(t) = [tex]-\frac{t}{20}e^{-\frac{t}{20} } + e^{-\frac{t}{20} }[/tex]
Now,
For minimum velocity, Put [tex]\frac{d}{dt}(v(t)) = 0[/tex]
Now,
[tex]\frac{d}{dt}[v(t)] = \frac{d}{dt} [ -\frac{t}{20}e^{-\frac{t}{20} } + e^{-\frac{t}{20} } ][/tex]
= [tex]-\frac{2}{20} e^{-\frac{t}{20} } + \frac{t}{400} e^{-\frac{t}{20} }[/tex]
Now,
Put [tex]\frac{d}{dt}(v(t)) = 0[/tex], we get
[tex]-\frac{2}{20} = - \frac{t}{400}[/tex]
⇒t = 40
Now,
Check that the point is minimum or maximum
Calculate [tex]\frac{d^{2} }{dt^{2} } [v(t)][/tex]
Now,
[tex]\frac{d^{2} }{dt^{2} } [v(t)][/tex] = [tex]\frac{d}{dt} [ -\frac{2}{20} e^{-\frac{t}{20} } + \frac{t}{400} e^{-\frac{t}{20} }][/tex]
= [tex]\frac{1}{400}e^{- \frac{t}{20} } [ 3 - \frac{t}{20}][/tex]
⇒[tex]\frac{d^{2} }{dt^{2} } [v(t)][/tex] = [tex]\frac{1}{400}e^{- \frac{t}{20} } [ 3 - \frac{t}{20}][/tex] > 0
∴ we get
t = 40 is point of minimum
So,
The minimum velocity be
v(40) = [tex]-\frac{40}{20}e^{-\frac{40}{20} } + e^{-\frac{40}{20} }[/tex]
= [tex]-2e^{-2 } + e^{-2 }[/tex]
= [tex]-e^{-2 }[/tex]
⇒v(40) = [tex]-e^{-2 }[/tex] units
∴ we get
The minimum velocity of the particle = [tex]-e^{-2 }[/tex] units
