M4 Relative Velocity - Shortest Distance?

Hey, I'm really struggling with shortest distance between two moving objects in M4. I can't see an example on it in the heinemann book either. Could someone please work through the following questions or give me some pointers?

ex1B
1) P and Q are two aircraft. At twelve noon a position vector of P relative to O is (5i + 7j) and for Q (115i - 25j) The velocity vectors of P and Q ae (100i - 285j)km/h and (75i - 135j)km/h respectively. Calculate the closest distance between the two aircraft.

^^ I can do that question working out the time first, but is that necessary?

2) Two cars are travelling along two straight roads which cross at right angles. One car is 200m away and approaches at 20m/s. Another is 100 m away and approaching at a speed of 25m/s (so there velocities are perpendicular). Calculate the shortest distance between the cars in the subsequent motion.

Cheers,
Graham

Reply 1

Christophicus

18

rugbygraham

Hey, I'm really struggling with shortest distance between two moving objects in M4. I can't see an example on it in the heinemann book either. Could someone please work through the following questions or give me some pointers?

ex1B
1) P and Q are two aircraft. At twelve noon a position vector of P relative to O is (5i + 7j) and for Q (115i - 25j) The velocity vectors of P and Q ae (100i - 285j)km/h and (75i - 135j)km/h respectively. Calculate the closest distance between the two aircraft.

^^ I can do that question working out the time first, but is that necessary?

Rp = (5i+7j) + (100i-285j)t
Rq = (115i-25j) + (75i-135j)t

Then calculate pRq, square it and differentiate it using the chain rule, find t and bung it into |pRq|
I think that's the easiest method anyway.

Reply 2

Christophicus

18

rugbygraham

2) Two cars are travelling along two straight roads which cross at right angles. One car is 200m away and approaches at 20m/s. Another is 100 m away and approaching at a speed of 25m/s (so there velocities are perpendicular). Calculate the shortest distance between the cars in the subsequent motion.

Right, this can be done using a diagram. Draw a velocity triangle and calculate various angles between the relative velocity and an axis. Now draw a line AoBo indicating the initial displacement between the two objects. The minimum distance is the perpendicular distance between the relative velocity vector and the line AoBo.

Reply 3

meef cheese

10

Widowmaker

Rp = (5i+7j) + (100i-285j)t
Rq = (115i-25j) + (75i-135j)t

Then calculate pRq, square it and differentiate it using the chain rule, find t and bung it into |pRq|
I think that's the easiest method anyway.

I always use pRq.pVq = 0 and solve to find t, then as you say, put it into |pRq|. Gives you the same answer, just another method.

Reply 4

GoldenGrizzly

I think in the book they start with the differentiation method to illustrate the fact you can just find the dot of relative distance with the relative velocity. pRq.pVq = 0 is always the way to go IMO. (Which makes sense anyway since we're trying to find a value for time such that the relative distance is perpendicular to the path of the relative velocity.)
For the other question, I would change it to i and j format, it's the easiest way I think. And do the exact same thing using pRq.pVq = 0 etc... ?

M4 Relative Velocity - Shortest Distance?

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