A long distance runner starts at the beginning of a trail and runs at a rate of 4 miles per hour. two hours later, a cyclist sta rts at the beginning of the trail and travels at a rate of 14 miles per hour. what is the amount of time that the cyclist travels before overtaking the runner?
1 answer:
To solve this problem, let us first assign variables. Let
us say that:
A = runner
B = cyclist
d = distance
v = velocity
time = t
The time in which the cyclist overtakes the runner is the
time wherein the distance of the two is the same, that is:
dA = dB
We know that the formula for calculating distance is:
d = v t
therefore,
vA tA = vB tB
Further, we know that tA = tB + 2, therefore:
vA (tB + 2) = vB tB
4 (tB + 2) = 14 tB
4 tB + 8 = 14 tB
10 tB = 8
tB = 0.8 hours = 48 min
Therefore the cyclist overtakes the runner after 0.8
hours or 48 minutes.
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![\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ -1 &,& -3~) % (c,d) &&(~ -7 &,& -5~) \end{array} \\\\\\ d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ r=\sqrt{[-7-(-1)]^2+[-5-(-3)]^2}\implies r=\sqrt{(-7+1)^2+(-5+3)^2} \\\\\\ r=\sqrt{36+4}\implies r=\sqrt{40}\\\\ -------------------------------](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%5C%5C%0A%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%26%26x_2%26%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%26%28~%20-1%20%26%2C%26%20-3~%29%20%0A%25%20%20%28c%2Cd%29%0A%26%26%28~%20-7%20%26%2C%26%20-5~%29%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ar%3D%5Csqrt%7B%5B-7-%28-1%29%5D%5E2%2B%5B-5-%28-3%29%5D%5E2%7D%5Cimplies%20r%3D%5Csqrt%7B%28-7%2B1%29%5E2%2B%28-5%2B3%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ar%3D%5Csqrt%7B36%2B4%7D%5Cimplies%20r%3D%5Csqrt%7B40%7D%5C%5C%5C%5C%0A-------------------------------)
![\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-1}{ h},\stackrel{-3}{ k})\qquad \qquad radius=\stackrel{\sqrt{40}}{ r} \\\\\\\ [x-(-1)]^2+[y-(-3)]^2=(\sqrt{40})^2\implies (x+1)^2+(y+3)^2=40](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%0A%28x-%20h%29%5E2%2B%28y-%20k%29%5E2%3D%20r%5E2%0A%5Cqquad%20%0Acenter~~%28%5Cstackrel%7B-1%7D%7B%20h%7D%2C%5Cstackrel%7B-3%7D%7B%20k%7D%29%5Cqquad%20%5Cqquad%20%0Aradius%3D%5Cstackrel%7B%5Csqrt%7B40%7D%7D%7B%20r%7D%0A%5C%5C%5C%5C%5C%5C%5C%0A%5Bx-%28-1%29%5D%5E2%2B%5By-%28-3%29%5D%5E2%3D%28%5Csqrt%7B40%7D%29%5E2%5Cimplies%20%28x%2B1%29%5E2%2B%28y%2B3%29%5E2%3D40)
