Answer:
x = 2.26 miles from the shore P i.e. point at which boat be landed.
Explanation:
x is the distance from point of landing i.e R and P
So, she cover a distance [tex]= BR = \sqrt{x^2+4}[/tex]miles
She rows with a speed of 3 miles per hour.
So, time taken for rowing[tex] = t_r = \frac{\sqrt{x^2+4}}{3}[/tex]
She walks a distance = RQ = 14-x
She walks with a speed of 4 miles per hour.
So, time taken by the woman for walking \
[tex]= t_w = \frac{14-x}{4}[/tex]
Thus, total time taken
[tex]= T = tr + tw =\frac{\sqrt{x^2+4}}{3} + \frac{14-x}{x}[/tex]
For total time to be least, T ' = 0
T '[tex] = \frac{x}{3\sqrt{x^2+4}} - \frac{1}{4}[/tex]
[tex]4x = 3\sqrt{x^2+4}[/tex]
[tex]16x^2 - 9x^2 = 36[/tex]
x = 2.26 miles from the shore P
Answer:
The boat be landed 2.26 miles away in order to arrive at a town.
Explanation:
Given data:
Distance of island from nearest point P is, [tex]d_{1}= 2\;\rm miles[/tex].
Speed of rowing boat is, [tex]v=3 \;\rm mi/h[/tex].
Speed of walking is, [tex]v' = 4 \;\rm mi/h[/tex].
Distance between the town to shore is, [tex]d_{2}=14 \;\rm miles[/tex].
Let x be the distance between the point of landing (say Z) and P. Then, distance covered by woman is,
[tex]d'=\sqrt{x^{2}+d^{2}_{1}}\\d'=\sqrt{x^{2}+2^{2}}\\d'=\sqrt{x^{2}+4}[/tex]
Time taken to cover the distance d' is,
[tex]v= \dfrac{d'}{t}\\3= \dfrac{\sqrt{x^{2}+4} }{t}\\t= \dfrac{\sqrt{x^{2}+4} }{3}[/tex]
Walking distance (d'') is,
[tex]d'' = d_{2}-x\\d'' = 14 -x[/tex]
Time taken to cover the walking distance is,
[tex]v'=\dfrac{d''}{t'} \\t'=\dfrac{d''}{v'} \\t'=\dfrac{14-x }{4}[/tex]
Let T be total time taken. Then,
[tex]T =t+t'\\T=\dfrac{\sqrt{x^{2}+4} }{3} + \dfrac{14-x}{4}[/tex]
For least time, T' = 0.
[tex]T'=0\\\dfrac{x }{3\sqrt{x^{2}+4}} - \dfrac{1}{4} =0\\4x =3\sqrt{x^{2}+4}\\16x^{2}=9(x^{2}+4)\\16x^{2}=(9x^{2}+36)\\x=\sqrt{\dfrac{36}{7} }\\x = 2.26 \;\rm miles[/tex]
Thus, the boat be landed 2.26 miles away in order to arrive at a town.
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