Respuesta :

Answer:

Step-by-step explanation:

Given that acceleration of an object is

[tex]\frac{dv}{dt} =-2v\\\frac{dv}{v} =-2dt\\ln v = -2t+C\\[/tex]

is the solution to the differential equation

Since v(0) =7

we get ln 7 = C

Hence [tex]lnv = -2t+ln 7\\v=7e^{-2t}[/tex]

since velocity is rate of change of distance s we have

[tex]v=\frac{ds}{dt} =7e^{-2t}\\s= [tex]s(t) =\frac{-7}{2} (e^{-2t})+C)[[/tex]

substitute t=0 and s=0

[tex]C=7/2[/tex]

So solution for distance is

[tex]s(t) =\frac{-7}{2} (e^{-2t}-1)[/tex]

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