A man and a woman start from the same point. The man walks S60°E at 1.3 m/sec. The woman
walks south at 1.6 m/sec. At what rate is the distance between the man and woman changing
after 35 minutes?​

Respuesta :

Answer:

The distance between the man and woman is changing at the rate of 1.5 m/sec.

Step-by-step explanation:

Speed = [tex]\frac{distance}{time}[/tex]

⇒ distance = speed x time

After 35 minutes (2100 seconds) ,

the man has walked a distance = 1.3 x 2100

                                                   = 2730 m

After 35 minutes, the woman has walked a distance = 1.6 x 2100

                                                    = 3360 m

The sketch of the displacement of the man and woman forms a triangle with an included angle. Applying the cosine rule, we have:

[tex]c^{2}[/tex] = [tex]a^{2}[/tex] + [tex]b^{2}[/tex] - 2abCos θ

   = [tex]3360^{2}[/tex] + [tex]2730^{2}[/tex] - 2 x 3360 x 2730 x Cos [tex]60^{o}[/tex]

   = 11289600 + 7452900 - 18345600(0.5)

   = 18742500 - 9172800

  = 9569700

c = [tex]\sqrt{9569700}[/tex]

  = 3093.5 m

The distance between the man and woman at 35 minutes is 3093.5 m.

The distance between the man and woman is changing at the rate = [tex]\frac{3093.5}{2100}[/tex]

                              = 1.4731

                              = 1.5 m/sec

The distance between the man and woman is changing at the rate of 1.5 m/sec.

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