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
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?
1 answer
1 answer