A few Specialist Problems

[/0]

Position vector is

    [v(t) is always tangent to r(t)]

When , .

The unit vector is

[Flaming_Arrow]

r(t) =6ti + (t^2+4)j
r'(t) = 6i + 2tj
r'(4) = 6i + 8j
|r'(4)| = rt[36 + 64]
          = 10

unit vector = r'(4) / |r'(4)|

(6i + 8j) / 10

[GerrySly]

Thanks for that guys, appreciate the help



I'm having trouble with part c. I have no idea how to find out the time when they are "nearest" each other.

[TrueTears]

There's 2 ways depending on whatever floats your boat.

1. The magnitude of |PQ| can be used. Find the minimum under the square root and thus you found the minimum t at which this occurs. [This is utilizing the result from b) ii)]

2. You can also dot the velocity vectors of OP and PQ and find when it equals 0, you must dot the velocity vectors and not the displacement because they give the direction of motion. [Since two particles are closest when they are perpendicular to each other, ie a straight line, notice this would only work if the path they travel in is a straight line.] But for this question you don't know the velocity vector for time t thus you should go with option 1.

[Mao]

There's 2 ways depending on whatever floats your boat.

1. The magnitude of |PQ| can be used. Find the minimum under the square root and thus you found the minimum t at which this occurs. [This is utilizing the result from b) ii)]

2. You can also dot the velocity vectors of OP and PQ and find when it equals 0, you must dot the velocity vectors and not the displacement because they give the direction of motion. [Since two particles are closest when they are perpendicular to each other, ie a straight line, notice this would only work if the path they travel in is a straight line.] But for this question you don't know the velocity vector for time t thus you should go with option 1.

the first method is valid.

However, I'm not sure how you got the second method. Two objects do not have to have perpendicular velocity at the closest point, this question included. [The velocities here are constant, not dependent on time]

[GerrySly]

I'm having trouble with part e. I know we have to use the dot product and solve for the cosine then arccos it, but I'm not sure what two vectors to use. Do I take a vector one second after being hit and the position of the hole, 100i, just not sure.

Cheers

[kamil9876]

There's 2 ways depending on whatever floats your boat.

1. The magnitude of |PQ| can be used. Find the minimum under the square root and thus you found the minimum t at which this occurs. [This is utilizing the result from b) ii)]

2. You can also dot the velocity vectors of OP and PQ and find when it equals 0, you must dot the velocity vectors and not the displacement because they give the direction of motion. [Since two particles are closest when they are perpendicular to each other, ie a straight line, notice this would only work if the path they travel in is a straight line.] But for this question you don't know the velocity vector for time t thus you should go with option 1.

the first method is valid.

However, I'm not sure how you got the second method. Two objects do not have to have perpendicular velocity at the closest point, this question included. [The velocities here are constant, not dependent on time]

The second method is reminiscent of method used when you have a path(possible straight line) and want to find the minimum distance between a FIXED point and the path, except you would take the dot product of velocity and the vector connecting the 'vehicle' and fixed point. (although there possibly could be a lesser distance at the end points of the path, or you could have several of these perpendicular cases(but this is mostly rare). Completely analogous to finding minimum by setting gradient to zero, in fact it's the same if you think about it geometrically).

edit: actually this method can sometimes give you maximum distance. By drawing a graph you can see whether it is minimum, max or something else (again, completely analogous to setting derivative to zero when finding min/max of y=f(x)). So I'll just always go with method 1 unless it's something obvious like a striaght line or curve bending away from (convex to??) FIXED point in question.

[GerrySly]

[kamil9876]

turning these into vectors:

3i+1j and ai+bj.

The angle between these must be and the magnitude of both must be equal. Using the dot product:




but (from equating magnitudes).





and use the the info that a and b are positive.

[Mao]

Another method:

OP = 3i + j
OQ = ai + bj
PQ = (a-3)i + (b-1)j


hence (1)

(2)

equating them gives , substituting back into (1) will yield the same answer as kamil

[Mao]

There's 2 ways depending on whatever floats your boat.

1. The magnitude of |PQ| can be used. Find the minimum under the square root and thus you found the minimum t at which this occurs. [This is utilizing the result from b) ii)]

2. You can also dot the velocity vectors of OP and PQ and find when it equals 0, you must dot the velocity vectors and not the displacement because they give the direction of motion. [Since two particles are closest when they are perpendicular to each other, ie a straight line, notice this would only work if the path they travel in is a straight line.] But for this question you don't know the velocity vector for time t thus you should go with option 1.

the first method is valid.

However, I'm not sure how you got the second method. Two objects do not have to have perpendicular velocity at the closest point, this question included. [The velocities here are constant, not dependent on time]

The second method is reminiscent of method used when you have a path(possible straight line) and want to find the minimum distance between a FIXED point and the path, except you would take the dot product of velocity and the vector connecting the 'vehicle' and fixed point. (although there possibly could be a lesser distance at the end points of the path, or you could have several of these perpendicular cases(but this is mostly rare). Completely analogous to finding minimum by setting gradient to zero, in fact it's the same if you think about it geometrically).

edit: actually this method can sometimes give you maximum distance. By drawing a graph you can see whether it is minimum, max or something else (again, completely analogous to setting derivative to zero when finding min/max of y=f(x)). So I'll just always go with method 1 unless it's something obvious like a striaght line or curve bending away from (convex to??) FIXED point in question.

yes yes yes, I see now. For moving objects P and Q, the dot product is , where the velocity must be non-zero, and either P and Q are travelling parallel to each other, or they intersect.

[bigtick]

OR in polar form.
p=rt(10)cis(artan1/3)
q=rt(10)cis(artan1/3 + pi/3)

.:a=rt(10)cos(artan1/3 + pi/3) and b=rt(10)sin(artan1/3 + pi/3).
Use compound angle formulae to evaluate.

[GerrySly]

Thanks guys Always good to see multiple approaches at questions



I'm not sure how the book got . I keep getting instead of what they want. The only way I could incorporate the speeds was to find the acceleration and then to find F=ma and solve.

[Flaming_Arrow]

u= 50/3 v = 20/3 t = 5

v = u + at
20/3 = 50/3 + 5a
a= -2

f = 1000 * 2
= 2000N

[/0]

did you convert to m/s :O