Answer:
a
[tex]u = 10.11 \ m/s[/tex]
b
[tex]x = 44.996 \ m[/tex]
Explanation:
From the question we are told that
The velocity of the river due east is [tex]v = 1.150i \ m /s[/tex]
The velocity of the boat due north is [tex]v_1 = 10j \ m/ s[/tex]
The width of the river is [tex]d = 300 \ m[/tex]
Generally the velocity of the boat relative to the shore
[tex]u = \sqrt{v^2 + v_1^2}[/tex]
=> [tex]u = \sqrt{1.50 ^2 + 10^2}[/tex]
=> [tex]u = 10.11 \ m/s[/tex]
Generally direction of the boat relative to the shore is mathematically represented as
[tex]tan \theta = \frac{v}{v_1}[/tex]
=> [tex]\theta = tan^{-1}[ \frac{v}{v_1}][/tex]
=> [tex]\theta = tan^{-1}[ \frac{1.5}{10}][/tex]
=> [tex]\theta = 8.53^o[/tex]
Generally the distance covered by the boat in the direction of flow of the river before reaching the north shore is mathematically represented as
[tex]x = d* tan(\theta )[/tex]
=> [tex]x = 300 * tan(8.53)[/tex]
=> [tex]x = 44.996 \ m[/tex]
(a) The velocity of the boat relative to the shore is 10.11 m/s.
(b) The distance traveled by the boat when it reaches the north shore is 303.3 m.
The given parameters;
(a) The velocity of the boat relative to the shore is calculated by applying Pythagoras theorem as shown below;
[tex]v = \sqrt{1.5^2 + 10^2} \\\\v = 10 .11 \ m/s[/tex]
(b) Let the time taken for the boat to reach the north shore = t
[tex]distance = speed \times time \\\\c^2 = a^2 + b^2\\\\[/tex]
[tex](10.11t)^2 = (300)^2 + (1.5t)^2\\\\102.21t^2 = 90,000 + 2.25t^2\\\\99.96t^2 = 90,000\\\\t^2 = \frac{90,000}{99.96} \\\\t^2 = 900\\\\t = \sqrt{900} \\\\t = 30 \ s[/tex]
The distance traveled by the boat when it reaches the north shore;
[tex]d = 10.11t\\\\d = 10.11 \times 30\\\\d = 303.3 \ m[/tex]
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