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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? $

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

02:53

Wen Z.

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 9

Related Rates

Derivatives

Differentiation

Missouri State University

University of Nottingham

Boston College

Lectures

03:09

In mathematics, precalculu…

31:55

In mathematics, a function…

06:05

At noon, ship $ A $ is $ 1…

04:55

Ships A and B leave port t…

07:05

Two boats leave a port at …

07:45

A boat leaves a dock at 2:…

07:53

Distance between two ships…

03:48

Marine Transportation A sh…

00:48

Boats A and B leave the sa…

06:42

NAVIGATION A ship lea…

01:46

A ship traveling east at $…

07:07

Ships Two ships are steami…

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.

Numerade Educator

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