Calculating Boat Speed in 2-D Relative Motion

In summary, the boat will have a speed of 4.54 km/h with respect to the Earth when it is heading south of east.
  • #36
R = (V_0tan14)/(1.2(9.8))/(V_0cos14)^2)
 
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  • #37
ok, now simplfy that equation and what do you get?
 
  • #38
R = V_0tan14/1 x 1/2(9.8)/(V_0cos14)^2

1.22V_0/(V_0cos14)^2

R = 1.30V_0
 
  • #39
check you math, you did something wrong.
you should re-work it and get this
R=(((V_0)^2)/9.8)(sin14cos14)

now, do you see how this equation is important?
 
  • #40
oh i see what I did wrong i forgot to flip the second equation when multiplying division.

actually wait how did you get that.

R = V_0tan14 x (V_0cos14)^2/1/2(9.8)...

where did 1/2 go
 
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  • #41
i wrote it out in microsoft equation to make it easier to understand

attachment.php?attachmentid=7776&stc=1&d=1158646032.jpg


the sine14cos14 becomes sin2(14) by the double angle forumla.

i got to go jogging for the night, so i'll be back in about a hour. but do you what this eqaution allows you to solve for?
 

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  • #42
the link said its invalid.
 
  • #43
I would say it will solve for the initial velocity.
 
  • #44
okay if i solve it correct V_0 = 22.61
 
  • #45
yes it will solve for initial velocity.
when i did the calculation for it and got 15.83 m/s

let me see if i can get the picture to work.
 
  • #46
BunDa4Th said:
the link said its invalid.
hmmm, the link works for me. if it still doesn't work let me know and i can e-mail it to you if you want.
 
  • #47
  • #48
i sent it to you. hopefull it will work and help.
 
  • #49
it works and i understand it a bit.

Im still a bit confuse how you get sin2(theta) and what does R equal to? or is it just there?
 
  • #50
well, the sin2(theta) came from the sin(theta)cos(theta); it's a trig idenity.
sin(x)cos(x)=sin2(x)

in the final equation we get, R is the range, or the distance travled in the x direction. let's go back to the example i used last time. if i threw a ball and it landed 10 meters away from me, then the range of the ball is 10 meters. range is just another way to say the distance that was travled in the x direction. in the motorcycles case, it needs to just a distance of 12 meters, the smallest range needed is 12 meters.
using algebra you can solve the equation for V_0. in other words, given an initial angle and distance travled (or required to travle) you can calculate the velocity needed.
 
  • #51
Oh man you are such a big help in solving this. I now understand it much better. Thanks very much for the time on helping me solve this equation step by step. Again I Thank you
 
  • #52
You're very welcome. :)
 

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