Respuesta :
Answer:
(a) 5.89 ft/s
(b) x= 1.77 ft
Explanation:
Given that the acceleration of the particle, a=-k/x
As [tex]a=v\frac{dv}{dx}[/tex] , so
[tex]v\frac{dv}{dx}=-\frac{k}{x} \\\\\Rightarrow vdv=-k\frac{dx}{x} \\\\[/tex]
On integrating both sides, we have
[tex]\frac{v^2}{2}=-k\ln(x) + C\cdots(i)[/tex]
where C is a constant.
At x=0.6 ft, v=15 ft/s
From equation (i)
[tex]\frac{15^2}{2}=-k\ln(0.6) + C \\\\\Rightarrow C=\frac{225}{2} + k\ln(0.6) \cdots(ii)[/tex]
Similarly, at x=1.2 ft, v=9 ft/s
[tex]\frac{9^2}{2}=-k\ln(1.2) + C \\\\\Rightarrow \frac{81}{2} = -k\ln(1.2) + \frac{225}{2} + k\ln(0.6) \\\\\Rightarrow k(\ln(1.2) - \ln(0.6))= \frac{225}{2}-\frac{81}{2} \\\\\Rightarrow k\ln(1.2/0.6)= 72 \\\\\Rightarrow k = 72 / ln(2)=103.87[/tex]
From equation (ii),
[tex]C=\frac{225}{2} + 103.87 \times \ln(0.6) \\\\\Rightarrow C=112.5-53.06=59.44[/tex]
Putting the value of k and C in the equation (i), we have
[tex]\frac{v^2}{2}=-103.87\ln(x) + 59.44 \\\\\Rightarrow v^2=-207.74\ln(x) + 118.88 \\\\\Rightarrow v = \sqrt{-207.74\ln(x) + 118.88}[/tex]
(a) At x=1.5 ft, the velocity of the particle is
[tex]\Rightarrow v = \sqrt{-207.74\ln(1.5) + 118.88} \\\\[/tex]
[tex]\Rightarrow v = 5.89[/tex] ft/s
At x=1.5 ft, the velocity of the particle is 5.89 ft/s.
(b) For v=0, we have
[tex]\Rightarrow 0 = \sqrt{-207.74\ln(x) + 118.88} \\\\\Rightarrow \ln(x) = -118.88/-207.74=0.572 \\\\\Rightarrow x= e^{0.572} \\\\[/tex]
[tex]\Rightarrow x= 1.77[/tex] ft
At x= 1.77 ft the velocity of the particle is zero.
