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nikklg [1K]
4 years ago
7

The falling object in Example 2 satisfies the initial value problem dv/dt =9.8−(v/5), v(0) =0. (a) Find the time that must elapse

for the object to reach 98% of its limiting velocity. (b) How far does the object fall in the time found in part (a)?
Physics
1 answer:
UNO [17]4 years ago
5 0

Answer:

a. t=19.56 s

b.d=718.34

Explanation:

The solution to the differential equation

\dfrac{dv}{dt}=9.8-\dfrac{v}{5}

is the exponential function

v(t)=ce^{-0.2t}+49

and we find c from the initial condition v(0)=0:

0=ce^{-0.2*0}+49\\\\0=c+49\\\\c=-49

Therefore, we have

v(t)=-49e^{-0.2t}+49

\boxed{ v(t)=49(1-e^{-0.2t})}

Part A:

The maximum velocity that the object can reach is 49 (which the maximum value v(t) can have).

Now, 98% of 49 is 48.02; therefore,

48.02=49(1-e^{-0.2t})

0.98=1-e^{-0.2t}

e^{-0.2t}=0.02

\boxed{t=19.56 s}

Part B:

The distance traveled is the integral of the speed:

d=\int_0^{19.56}v(t)*dt

d=\int^{19.56}_0 {49(1-e^{-0.2t})} \, dt

d=49[t+5e^{-0.2t}]_0^{19.56}

\boxed{d=718.34}

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Given that:

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Answer:

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