SOLUTION: help please One train leaves a station at noon heading east at a rate of 60.0 km/h. A second train leaves the same station at 2PM heading north at a rate of 75.0 km/h. Find the

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Question 1163169: help please
One train leaves a station at noon heading east at a rate of 60.0 km/h. A second train leaves the same station at 2PM heading north at a rate of 75.0 km/h. Find the rate at which they are separating at 5PM.
Round to 3 significant digits
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Let t=0 (hours) be at 2pm. Then

The position of train A at t=0 is (120,0). (At 2pm, it has been traveling east at 60km/h.)
The position of train A at time t is (60t+120,0).

The position of train B at time t is (0,75t).

The distance between them at time t is

d+=+sqrt%28%2860t%2B120%29%5E2%2B%2875t%29%5E2%29

We need to find the rate at which that distance is increasing at 5pm (at t=3).

To avoid differentiating an ugly square root expression, I will use the square of the distance....



2d%2A%28dd%2Fdt%29+=+18450t%2B14400

We need to evaluate dd/dt at t=3; that means we need to know the distance d between the trains at t=3:

d%283%29+=+sqrt%28300%5E2%2B225%5E2%29+=+375

(Note that calculation is easy if you recognize that the legs 225 and 300 are scale models of a 3-4-5 right triangle, making the hypotenuse 375.)

So at t=3,

2d%2A%28dd%2Fdt%29+=+18450t%2B14400

2%28375%29%2A%28dd%2Fdt%29+=+18450%283%29%2B14400+=+69750

dd%2Fdt+=+69750%2F750+=+93

ANSWER: At 5pm, the two trains are separating at a rate of 93km/h.