Question 759069
two straight paths,inclined to one another at 60 degrees,interesct at a point o.A boy A is on one path 300m from O while a boy B is on the other path 400m from O.angle AOB=60 degrees.Simultaneously the boys begin to run towards O,A with speed 15 kmh-1 and B with speed 12 kmh-1 What is the shortest distance between the two boys?
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The speed of A = 250 meters/min
Speed of B = 200 m/min
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a & b = distance from O at time t (in minutes)
a = 300 - 250t
b = 400 - 200t
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d = distance between A & B at time t
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{{{d^2 = a^2 + b^2 - 2ab*cos(60)}}}
{{{d^2 = a^2 + b^2 - ab}}}
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{{{d^2 = 2500*(21t^2 - 60t + 52)}}}
Without calculus:
Finding the minimum for d is the same as finding the minimum for d^2
The min is the vertex of the parabola {{{d^2 = 2500*(21t^2 - 60t + 52)}}}
The vertex is on the LOS, Line of Symmetry t = -b/2a
t = 60/42 = 10/7 seconds
Sub for t as below
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Using the 1st derivative:
{{{d(t) = 50*(21t^2 - 60t + 52)^(1/2)}}}

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d'(t) = {{{50*(1/2)*(21t^2 - 60t + 52)^(-1/2)*(42t - 60)}}}
d'(t) = {{{150*(21t^2 - 60t + 52)^(-1/2)*(7t - 10)}}}
@ d'(t) = 0:
{{{150*(21t^2 - 60t + 52)^(-1/2)*(7t - 10) = 0}}}
{{{7t - 10 = 0}}}
t = 10/7 minutes
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d(10/7) = 151.186 meters