Vectors mechanics

Hi! Sorry to be annoying, but could someone please show me how to calculate courses set for closest approach/ true velocities given two situations where the true velocity is unknown/ courses set and time taken for a collision using i,j,k component form (without the use of geometry in a diagram?) Many thanks in advance :smile:

Reply 1

python03

Original post by marinacalder

do you have a question that asks for ^ that?

Reply 2

username2436711

Original post by leekn

do you have a question that asks for ^ that?

(edited 6 years ago)

any ideas?

Theoretically use i, j vectors... you have to draw a diagram, no way around that tbh :/

Reply 6

ghostwalker

Original post by marinacalder

any ideas?

"Ghost Walker" is a different user, who hasn't been active for 14 years.

Post #3

As leekn said, I,j components will cover it.

Vel. wind = vel ferry + vel. wind relative to ferry.

Similarly the trawler.

Since you only know the directions for the wind, and not the speed, your relative velocites will be of the form a(bI+cJ) for some a, unknown, and different for each equation. A total of 4 unknowns.

Post some working if you're stuck.

(edited 6 years ago)

Reply 7

RDKGames

Study Forum Helper

Geometric approach is much more straight-forward if you know how to do it properly...

Reply 8

username2436711

Original post by RDKGames

Geometric approach is much more straight-forward if you know how to do it properly...

I seem to be getting answers that are similar to the answers in the book, but are usually around +/- 0.5hm out...

I'm not very good at this at all :frown:

Reply 9

username2436711

Original post by marinacalder

I seem to be getting answers that are similar to the answers in the book, but are usually around +/- 0.5hm out...

I'm not very good at this at all :frown:

I've never been taught this and have no idea where to start...

Reply 10

RDKGames

Study Forum Helper

Original post by marinacalder

I've never been taught this and have no idea where to start...

For the shortest time to intercept, you want the velocity of the fielder relative to the ball be along the path of the distance between the batsman and the fielder.

(edited 6 years ago)

Reply 11

Physics Enemy

Original post by marinacalder

If you sketch it, all the action is in the neg (3rd) quadrant. Ball's path lies 10° to the x-axis. Fielder starts at (- 45cos25°, - 45sin25°), running at clockwise @ to his North line, aiming to intercept the ball.

r_b = - (15cos10°)t i - (15sin10°)t j
r_f = [(6sin@)t - 45cos25°] i + [(6cos@)t - 45sin25°] j

At interception: Equate i co-effs, j co-effs

i: 45cos25° - (6sin@)t = (15cos10)t
=> t(6sin@ + 15cos10°) = 45cos25°

j: 45sin25° - (6cos@)t = (15sin10°)t
=> t(6cos@ + 15sin10°) = 45sin25°

So tan25° = (6cos@ + 15sin10°)/(6sin@ + 15cos10°)
6(tan25°)sin@ + 15tan25°cos10° = 6cos@ + 15sin10°
=> (tan25°)sin@ - cos@ = 2.5(sin10° - tan25°cos10°)

(tan25°)sin@ - cos@ = Rsin(@ - a) = Rcos(a).sin@ - Rsin(a).cos@

R^2 = 1 + (tan25°)^2 = (sec25°)^2 => R = sec25°
tan(a) = 1/tan25° => a = 65°

So (sec25°)sin(@ - 65°) = 2.5(sin10° - tan25°cos10°)
=> sin(@ - 65°) = 2.5(sin10°cos25 - sin25°cos10°)
=> sin(@ - 65°) = - 2.5sin15°
=> @ - 65° = - 40.32°
=> @ = 24.7°, Fielder's path on N 24.7° E

For t, sub @ = 24.68° into i co-effs eqn:
t = (45cos25°)/(6sin24.68° + 15cos10°)
=> t = 2.3605, t = 2.4 s (1 d.p)

(edited 6 years ago)

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