The good ship bravery is 30 km due west of the good ship courageous the bravery sets on on a bearing of 30°?
at a speed of 25 km/h and the courageous sets out on a bearing of 345° at a speed of 20 km/h will the ships collide
*need help drawing DIAGRAM!!*
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Let horizontal position of A = 0
horizontal position of C = 30
Horizontal velocity of A = 25cos 60
Horizontal velocity of C = 20cos75
Horizontal positions after t hours
A = 25tcos60
C = 30-20tcos75
For a collision A = C
25tcos60 = 30-20tcos75
t = 1.6972 hours
For the vertical situation
A = 1.6972*25cos30 = 36.7455 km
C = 1,6972*20cos15 = 32.7874
So they will miss by about 4 km
Answer No collision.
Make an x-y coord. axis .. but the pos y axis is North, and the neg y is South. Similarly, the pos x-axis is East and the neg x- is west.
Point C is the present position of the Courageous. It is the ORIGIN.
Point B is the present position of the Bravery. It is at (-30,0).
B is traveling at 25km/h on a course of 30 deg (theta)... this translates to a linear equation .. slope f the line of motion is tan theta
... m = tan 20 ... 0.36397 and the x intercept is -30
y - 0 = 0.36397(x - 30)
y = 0.36397x - 10.9191
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similarly the lune for C ... traveling at 20km/h
m = tan345 = -0.26795 and the y intercept is 0
y = -0.26795x
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intersection of the 2 lines ...
0.36397x - 10.9191 = -0.26795x
x = 17.279 ... y = -4.62997
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but ... are these at the same time? meaning a collision occurs
... Pythag. Theorem .. dist. B moves is 13.537 km at 25k/h
t = 0.5415 hour
again for dist C moves ... d = 17.888557 km
t = 0.89443 hour .. not same time, so no collision