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Numerade Educator

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Problem 23 Medium Difficulty

At noon, ship $ A $ is $ 100 km $ west of ship $ B. $ Ship $ A $ is sailing south at $ 35 km/h $ and ship $ B $ is sailing north at $ 25 km/h. $ How fast is the distance between the ships changing at $ 4:00 PM? $

Answer

$$55.4 \mathrm{km} / \mathrm{h}$$

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Video Transcript

okay. Using the distance formula to figure out the distance between the two ships at time t we know we have 100 minus zero square. Plus, this is where variables come in 25 t post 35 t squared, which gives us DDT is the square root 10,000 plus 3600 t squared, which now tells us that deep prime of tea is gonna be one over two square root of 10 1000 plus 3600 t squared which simplifies to 1 80 t divided by the square root of 25 plus 90 squared, which tells us that deep REM of force notices just plugging in four is gonna be 1 80 times four divided by squirt of 25 plus nine times for squared, which is gonna be 55.38 kilometers per hour.