Ship A is located 4.0 km north and 2.5 km east of ship B. Ship A has a velocity of 22 km/h toward the south, and ship B has a velocity of 40 km/h in a direction 37° north of east.
(a) What is the velocity of A relative to B in unit-vector notation with ˆi toward the east?
(b) Write an expression (in terms of ˆi and ˆj) for the position of A relative to B as a function of t, where t ???? 0 when the ships are in the positions described above.
(c) At what time is the separation between the ships least?
(d) What is that least separation?

Question

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Respuesta :

lcmendozaf lcmendozaf · Answer 1 · 2019-10-09T03:39:24+02:00

Answer:

a) [tex]V_{A/B}=31.95(-i) + 46.07(-j) km/h[/tex]

b) [tex]r_{A/B}(t)=(2.5-31.95t)(i)+(4-46.07)(j)km[/tex] where t is in hours

c) t=0.084h

d) d=0.22km

Explanation:

We know that:

[tex]V_{A/B}=V_A-V_B=(0,-22)-(40*cos(37),40*sin(37))[/tex]

[tex]V_{A/B}=(-31.95,-46.07)km/h[/tex]

For the position:

[tex]r_{A/B}(t)=ro_{A/B}+V_{A/B}*t[/tex]

[tex]r_{A/B}(t)=(2.5,4)+(-31.95,-46.07)*t=(2.5-31.95*t,4-46.07*t)km[/tex]

The separation is given by the module of their relative position:

[tex]d=\sqrt{(2.5-31.95*t)^2+(4-46.07*t)^2}[/tex] To find the least distance, we have to derive and make equal 0 to find t:

[tex]d'=\frac{2*(-31.95)*(2.5-31.95*t)+2*(-46.07)*(4-46.07*t)}{2*\sqrt{(2.5-31.95*t)^2+(4-46.07*t)^2}} =0[/tex] Solving for t:

t = 0.084h

Evaluating d when t=0.084 we get:

d=0.22km