Calculus

1)A north south highway, intersects an east west highway, at point P. A vehicle process point P @ 1pm, traveling east at a constant speed of 60km/h. At the same instant another vehiclist 5km north of P, travelling south at 80km/h, find the time, when the 2 vehicles are closest to each other and the distance between them at this time.

The answer is 1:02pm and distance of 3km, but idk how to get this can you please show the steps so I can understand this question well?

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  1. Make a sketch ...
    Place the first vehicle A at the origin P, showing the position of the first vehicle
    draw a short line east of P and label it A2

    Place the second vehicle B at a distance of 5 above P
    showing the initial position of the second vehicle
    draw line downwards from B towards P, label it B2

    At a time of t hours after 1:00,
    AA2 = 60t km
    BB2 = 80t km

    join A2 andB2 call it D
    D is the distance between the two endpoints of the cars.
    You now have a right-angled triangle with sides
    D, 60t, and 5-80t

    D^2 = (60t)^2 + (5-80t)^2
    = 3600t^2 + 25 - 800t + 6400t^2
    = 10000t^2 - 800t + 25

    when they are closest d(D)/dt = 0
    2D d(d)/dt = 20000t - 800
    d(D)/dt = (10000t - 400)/D
    = 0
    10000t = 400
    t = 400/10000 = .04 hrs

    .04 hrs = .04(60) or 2.4 minutes

    so they are closest at 1:00 + 0:02.4
    = 1:02.4 pm

    when t = .04
    D^2 = 10000(.04)^2 - 800(.04) + 25
    = 9
    D = √9 = 3

    So the closest distance between them is 3 km at 1:02.4 pm

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