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A falling object satisfies the initial value problem dv dt = 9.8 − v 5 , v(0) = 0 where v is the velocity in meters per second. (a) Find the time that must elapse for the object to reach 95% of its limiting velocity. (Round your answer to two decimal places.) s (b) How far does the object fall in the time found in part (a)? (Round your answer to two decimal places.) m Additional Materials

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Answer:

a.  t [tex]\simeq[/tex] 14.98 sec

b.   x = 501.27 m

Explanation:

From the given information:

[tex]\dfrac{dv}{dt}=9.8-(\dfrac{v}{5 })[/tex]  and   [tex]v(0)=0[/tex]

[tex]\dfrac{dv}{dt}=\dfrac{49-v}{5 }[/tex]

[tex]\dfrac{dv}{49-v}=\dfrac{dt}{5 }[/tex]

Taking  Integral of  both sides

[tex]- ln(49-v) = \dfrac{t}{5} + C[/tex]  

at t=0 we have v=0

This implies that

[tex]- ln(49-0) = \dfrac{0}{5} + C[/tex]

[tex]C= - ln(49)[/tex]

Thus:

[tex]\dfrac{t}{5} - In (49) = - In (49 -v) \\ \\ In(49) - \dfrac{t}{5} = In (49-v)[/tex]

[tex]49-v = e^{(-\frac{t}{5} +ln(49))}\\ \\ v = 49 - 49e^{(-\dfrac{t}{5})}[/tex]

The limiting velocity when the time is infinite is :

95% of 49 = 46.55

∴

[tex]0.05= e^{(-\dfrac{t}{5})}[/tex]

[tex]\dfrac{t}{5}= In(\dfrac{1}{0.05})[/tex]

[tex]\dfrac{t}{5}=2.9957[/tex]

t = 5 × 2.9957

t [tex]\simeq[/tex] 14.98 sec

b.) [tex]v = 49 - 49e^{(-\dfrac{t}{5})}[/tex]

[tex]v = \dfrac{dx}{dt}=49 - 49e^{(-\dfrac{t}{5})}[/tex]

[tex]dx=(49 - 49e^{(-\frac{t}{5})}) \ dt[/tex]

Taking integral of both sides.

[tex]x = 49t + 245 e^{(\frac{-t}{5})} +C[/tex]

 at time t = 0 , distance x traveled = 0

∴

C= - 245

Therefore

[tex]x = 49t + 245 e^{(\frac{-t}{5})} -245[/tex]

replacing the value of t = 14.98

[tex]x = 49(14.98) + 245 e^{(\frac{-14.98}{5})} -245[/tex]

x = 501.27 m

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