Nearest Distance between 2 boats

[zimbabwe]

Ship A is 10 km due west of ship B. Ship A is heading directly north at a speed of 30km/hr, whilst ship B is heading in a direction 60 degress west of norht at a speed of 20km/hr.

a) Deteremine the magnitude and direction of the velocity of ship B relative to ship A.

b) What will be their distance of closest approach?


My Attempt for the first part


B Relative to A, first into compentents, so sin(60)*20=17.3 and cos(60)*20=10
Then relative to A
would be 10 km west and 30-17.3= 12.67 km south

however the actual answer is Velocity is 26.5 km/h in a direction South 41 degrees West.

How was i supposed to do it, and how do i start part b.

 

[Alewhey]

Ship A is 10 km due west of ship B. Ship A is heading directly north at a speed of 30km/hr, whilst ship B is heading in a direction 60 degress west of norht at a speed of 20km/hr.

a) Deteremine the magnitude and direction of the velocity of ship B relative to ship A.

b) What will be their distance of closest approach?


My Attempt for the first part


B Relative to A, first into compentents, so sin(60)*20=17.3 and cos(60)*20=10
Then relative to A
would be 10 km west and 30-17.3= 12.67 km south

however the actual answer is Velocity is 26.5 km/h in a direction South 41 degrees West.

How was i supposed to do it, and how do i start part b.

For part a), you have simply got the components confused. Sin() gives the 'west' component not 'south', and vice versa. Try drawing the triangle to see your mistake (sin = opp/hyp, etc).

For part b), now that you have the relative velocity vector you can set up t-dependent vector R giving the relative position of the boats. Now the problem is simply a matter of finding the shortest distance between a point and a line. One way to do it is to construct a line which goes through the origin and whose dot-product with R is zero (therefore orthogonal to R). Then evaluate the point of intersection of the two lines and take the magnitude of the separation.

 

[zimbabwe]

Right i see part the mistake, so now i have 20 km/hr south and 17.3km/hr west. Find the hypontuse of this I get 26.4 km/hr, ok i got it. Thank you

b)i don't see it, by vector r does it mean, r=(10,0)+t(-17.3,-20)

 

[Alewhey]

Right i see part the mistake, so now i have 20 km/hr south and 17.3km/hr west. Find the hypontuse of this I get 26.4 km/hr, ok i got it. Thank you

b)i don't see it, by vector r does it mean, r=(10,0)+t(-17.3,-20)

Right, R gives the equation of a line in the x-y plane, with direction (-17.3,-20). So a vector perpendicular to the line could be

L = m(20,-17.3)

With m a parameter which can take any value (similarly to t).

We then find the points (x,y) where the two lines cross:

x = 20m = 10 -17.3t
y = -17.3m = -20t

Now solve the simultaneous equations => t=0.25 etc

A quicker way to do it is to spot that the point we want occurs at

R (dot) [direction of line] = 0

i.e. 10*-17.3 + t(-17.3^2 + -20^2) = 0

=> t = 0.25 etc

A third way to do it is to simply take the modulus of R and minimise it w.r.t. t. Takes a bit more algebra but requires less geometrical thinking.