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    The question is:

    Consider two particles, A and B, with positions, relative to a stationary observer, given by rA = ti - cos(2t) j - 3k, rB = sin(t)i + (3 + 2t)j + 7k

    Using vector algebra find, if any exist, all the times t such that the stationary observer would see A and B having the same velocity.

    I'm guessing its something to do with a Galilean transformation but I have no idea where to begin.

    Edit: managed to figure out - take derivatives of both position vectors wrto t, equate and solve for t. I obtained t=arcos(0) and t=arcsin(1/4).

    Now stuck on the following:
    Find t such that the stationary observer would see A is moving in a direction perpendicular to the motion of B.
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    (Original post by TajwarC)
    The question is:

    Consider two particles, A and B, with positions, relative to a stationary observer, given by rA = ti - cos(2t) j - 3k, rB = sin(t)i + (3 + 2t)j + 7k

    Using vector algebra find, if any exist, all the times t such that the stationary observer would see A and B having the same velocity.

    I'm guessing its something to do with a Galilean transformation but I have no idea where to begin.

    Edit: managed to figure out - take derivatives of both position vectors wrto t, equate and solve for t. I obtained t=arcos(0) and t=arcsin(1/4).

    Now stuck on the following:
    Find t such that the stationary observer would see A is moving in a direction perpendicular to the motion of B.
    Is this something as simple as the dot product of the two velocity vectors must equal zero?

    I hesitated to respond, in case the situation is more complicated than the standard A-level, and observer literally meaning that - view from a point . But from your solution to the first part, it looks to be standard 3D vector in a fixed frame of reference with the "observer" seeing the lot.

    Edit: Looking at your solution to the first part, with that method I got cos(t)=1 AND sin(t)=1/2 which has no solution.
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    (Original post by ghostwalker)
    Is this something as simple as the dot product of the two velocity vectors must equal zero?

    I hestitated to respond, in case the situation is more complicated than the standard A-level, and observer literally meaning that - view from a point . But from your solution to the first part, it looks to be standard 3D vector in a fixed frame of reference with the "observer" seeing the lot.
    Yes, this is exactly what I ended up doing. I calculated the dot product, set it equal to 0 and attempted to find t. In this case, I found that no such time exists.

    This is actually from a Dynamics and Relativity first year Math Undergraduate course.

    I'm now stuck with trying to find t when A is accelerating in (or opposite to) the direction that B is travelling. I've found the second derivative of the position vector of A but struggling to equate that to the direction to B.
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    (Original post by TajwarC)
    Yes, this is exactly what I ended up doing. I calculated the dot product, set it equal to 0 and attempted to find t. In this case, I found that no such time exists.
    I get cos(t) = 0 or sin(t) = -1/8 for when the velocities are perpendicular.

    Also, see "edit" in my previous post.

    This is actually from a Dynamics and Relativity first year Math Undergraduate course.

    I'm now stuck with trying to find t when A is accelerating in (or opposite to) the direction that B is travelling. I've found the second derivative of the position vector of A but struggling to equate that to the direction to B.
    If it's in the same direction then you'd have acceleration of A is a scalar multple of the velocity of B.
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    (Original post by ghostwalker)
    I get cos(t) = 0 or sin(t) = -1/8 for when the velocities are perpendicular.

    Also, see "edit" in my previous post.



    If it's in the same direction then you'd have acceleration of A is a scalar multple of the velocity of B.
    Made an arithmetic error for first part, got the same as you in your edit. Next part, I have the dot product as tsint-3cos2t-2tcos2t-21=0. Can't seem to reduce this down to what you got..

    (Thanks for the help thus far)
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    (Original post by TajwarC)
    Made an arithmetic error for first part, got the same as you in your edit. Next part, I have the dot product as tsint-3cos2t-2tcos2t-21=0. Can't seem to reduce this down to what you got..

    (Thanks for the help thus far)
    You've used the dot product on the position vectors rather than the velocity vectors.
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    (Original post by ghostwalker)
    You've used the dot product on the position vectors rather than the velocity vectors.
    Silly me, didn't notice it said "motion of B". For the last part, I found the acceleration is not a scalar multiple of the velocity of B, i.e.

    Second derivative of A is not a multiple of the derivative of B. Does this imply that no such time exists where the acceleration of A is in or opposite to the direction B is travelling?
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    (Original post by TajwarC)
    Silly me, didn't notice it said "motion of B". For the last part, I found the acceleration is not a scalar multiple of the velocity of B, i.e.

    Second derivative of A is not a multiple of the derivative of B. Does this imply that no such time exists where the acceleration of A is in or opposite to the direction B is travelling?
    It would do if this was true for all t, however I get cos(t)=0 as a solution.

    Last post for today.
 
 
 
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