Answer:
a. [tex]t=19.56 s[/tex]
b.[tex]d=718.34[/tex]
Explanation:
The solution to the differential equation
[tex]\dfrac{dv}{dt}=9.8-\dfrac{v}{5}[/tex]
is the exponential function
[tex]v(t)=ce^{-0.2t}+49[/tex]
and we find [tex]c[/tex] from the initial condition [tex]v(0)=0:[/tex]
[tex]0=ce^{-0.2*0}+49\\\\0=c+49\\\\c=-49[/tex]
Therefore, we have
[tex]v(t)=-49e^{-0.2t}+49[/tex]
[tex]\boxed{ v(t)=49(1-e^{-0.2t})}[/tex]
Part A:
The maximum velocity that the object can reach is 49 (which the maximum value [tex]v(t)[/tex] can have).
Now, 98% of 49 is 48.02; therefore,
[tex]48.02=49(1-e^{-0.2t})[/tex]
[tex]0.98=1-e^{-0.2t}[/tex]
[tex]e^{-0.2t}=0.02[/tex]
[tex]\boxed{t=19.56 s}[/tex]
Part B:
The distance traveled is the integral of the speed:
[tex]d=\int_0^{19.56}v(t)*dt[/tex]
[tex]d=\int^{19.56}_0 {49(1-e^{-0.2t})} \, dt[/tex]
[tex]d=49[t+5e^{-0.2t}]_0^{19.56}[/tex]
[tex]\boxed{d=718.34}[/tex]
