Question 999183
the x-coordinate of the tanker is equal to -18t + 162.
the y-coordinate of the tanker is 0.


the x-coordinate of the liner is 0.
the y-coordinate of the liner is 23t - 184.


t is the time in hours.


let x1 = -18t + 162
let y1 = 0


let x2 = 0
let y2 = 23t - 184


distance between the tanker and the liner at any given time is equal to:


sqrt((x1-x2)^2 + (y1-y2)^2)


this equation becomes:


d(t) = sqrt((-18t+162)^2 + (-23t+184)^2)


to graph this equation, we set y = d(t) and we set x = t.


the equation becomes:


y = sqrt((-18x+162)^2 + (-23x+184)^2)


if you simplify this, you get:


y = sqrt(853x^2 - 14296x + 60100)


that's a quadratic equation.


you can find the x-coordinate of the minimum point on that graph by using x = sqrt(-b/2a) formula.


the y-value of the minimum point on the graph is sqrt(f(-b/2a))


that point is shown on the graph as (8.38,14.8)


the points used to determine the distance equation were taken from a table created for t from t = 0 to t = 11 or 12.


the position of the tanker was (x1,y1)
the position of the liner was (x2,y2)


the distance equation became d(t) = sqrt((x1-x2)^2 + (y1-y2)^2)


for the tanker, y1 = 0
for the liner, x1 = 0


for the tanker, x1 = -18t + 162
for the liner, y2 = 23t + 184


these equations were derived by applying some logic.
for example:


when t = 9, the tanker was at (0,0)
the tanker went -18 units on the x-axis every hour.
therefore, when t = 0, the tanker has to be at x = 162 because:
f(t) = -18(0) + 162 = 162
when t = 9, the tanker has to be at x = 0.
f(t) = -18(9) + 162 = -162 + 162 = 0


simkilar logic was applied to the liner to make sure it was at y = 0 when t = 8.
g(t) = 23t - 184 was the equation.
when t = 8, 23t - 184 = 184 - 184 = 0


the equations were tested to make sure they were reliable.
once determined to be reliable, they were used to find the distance between the position of the tanker and the position of the liner at any given point in time.






that graph is shown below:


<img src = "http://theo.x10hosting.com/2015/102701.jpg" alt="$$$" </>


any questions, just send me an email.