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
