At noon, ship A is 130 km west of ship B. Ship A is sailing east at 30 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?
2 answers:
Answer:
36.1 km/h
Step-by-step explanation:
rate at which the distance is changing is given by relative velocity(vR)
Look at the diagram. THe diagram shows relative motion. From relative motion diagram, velocity vector diagram is drawn.
From velocity vector diagram, it can be seen that relative velocity or vR can be calculated using pythagoras theorem.
vR²= vA² + vB²
vR²= 30²+ 20²
vR= 36.1 km/h
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Step-by-step explanation:
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Answer:
The right response is "". A further solution is given below.
Step-by-step explanation:
The given expression is:
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On solving the above expression, we get
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On taking "3" as common, we get
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