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At noon, ship A is 100 km west of ship B. Ship A is sailing south at 40 km/h and ship B is sailing north at 20 km/h. How fast is the distance between the ships changing at 4:00 PM? (Round your answer to one decimal place.)

User Dan Fego
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1 Answer

2 votes

Answer:

The distance between the ships is changing at the rate of 113.9km/hr

Explanation:

Given that A is 100 km west of B. and A is sailing south at 40 km/h,

The coordinate of A is (0,-40t)

The coordinate of B is (100,30t)

Where t represent time from initial point to final point

Calculating the distance between both ships

d(t) = √((100 - 0)² + (30t - (-40t))

d(t) = √((100)² + (30t + 40t))

d(t) = √(10000 + (70t)²)

d(t) = √(10000 + 4900t²)

Differentiate with respect to t

d'(t) = ½(1/√(10000 + 4900t²))*9800t

d'(t) = 4900t/(√(10000 + 4900t²))

Simplify to lowest term

d'(t) = 4900t/(√(100(100+49t²)))

d'(t) = 4900t/(10√(100+49t²))

d'(t) = 490t/√(100+49t²)

Substitute 4 for t in the above

d'(4) = 490*4/√(100+49*4²)

d'(4) = 1960/√296

d'(4) = 113.9226859689428

d'(4) = 113.9

The distance between the ships is changing at the rate of 113.9km/hr

User Markus Kull
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