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    Vector Examples

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    anon1992
    S and B will collide at 30 seconds after noon at a position of 210i + 840j. To prevent collision, S changes its velocity to (7i + 30j) m/s after 15 seconds. At 30 seconds, S is at 210i + 1170j and B is at 210i + 840j. The distance between them is 330 meters.

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    Vector Examples

    Uploaded by

    anon1992
    195 views7 pages
    S and B will collide at 30 seconds after noon at a position of 210i + 840j. To prevent collision, S changes its velocity to (7i + 30j) m/s after 15 seconds. At 30 seconds, S is at 210i + 1170j and B is at 210i + 840j. The distance between them is 330 meters.

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    S and B will collide at 30 seconds after noon at a position of 210i + 840j. To prevent collision, S changes its velocity to (7i + 30j) m/s after 15 seconds. At 30 seconds, S is at 210i + 1170j and B is at 210i + 840j. The distance between them is 330 meters.

    Copyright:

    Attribution Non-Commercial (BY-NC)
    Download as PPT, PDF, TXT or read online from Scribd
    Download as ppt, pdf, or txt
    195 views7 pages

    Vector Examples

    Uploaded by

    anon1992
    S and B will collide at 30 seconds after noon at a position of 210i + 840j. To prevent collision, S changes its velocity to (7i + 30j) m/s after 15 seconds. At 30 seconds, S is at 210i + 1170j and B is at 210i + 840j. The distance between them is 330 meters.

    Copyright:

    Attribution Non-Commercial (BY-NC)
    Download as PPT, PDF, TXT or read online from Scribd
    Download as ppt, pdf, or txt
    of 7

    Question 11 Exercise 2D N

    At noon a ship S in 600m due north of an


    observation point O and a speedboat B is 120m due
    north of the same point. The ship S has constant
    velocity (7i + 8j) ms-1 and the speed boat has S
    constant velocity of (7i + 24j) ms-1.
    a) Write down the position vectors of S and B at the time t
    seconds after noon.
    position vector at time t = initial position + (velocity × t)
    B

    S rs  600 j   7i  8 j t Observation Point

    B rB  120 j   7i  24 j t
    Question 11 Exercise 2D N

    At noon a ship S in 600m due north of an


    observation point O and a speedboat B is 120m due
    north of the same point. The ship S has constant
    velocity (7i + 8j) ms-1 and the speed boat has S
    constant velocity of (7i + 24j) ms-1.
    a) Write down the position vectors of S and B at the time t
      7noon.
    rs  600 jafter
    seconds i  8 j t 
    r  120 j  7i  24 j t
    B 
    B

    S rs  600 j   7i  8 j t Observation Point

    B rB  120 j   7i  24 j t
    Question 11 Exercise 2D N

    At noon a ship S in 600m due north of an


    observation point O and a speedboat B is 120m due
    north of the same point. The ship S has constant
    velocity (7i + 8j) ms-1 and the speed boat has S
    constant velocity of (7i + 24j) ms-1.
    S B
    rs  600 j   7i  8 j t rB  120 j   7i  24 j t
    B
    b) Show that S and B will collide and find the time when
    this collision occurs and the position vector of the point of
    collision Observation Point

    For two objects to have a collision they must occupy the same space at the same time…

    So find a value for t such that:


    rs  rB S and B occupy the same
    600 j   7i  8 j t  120 j   7i  24 j t horizontal space for all t
    so just consider j

    i j
    Equating for i and j
     7t   7t 600   8 t  120   24  t
    t  30
    Question 11 Exercise 2D N

    At noon a ship S in 600m due north of an


    observation point O and a speedboat B is 120m due
    north of the same point. The ship S has constant
    velocity (7i + 8j) ms-1 and the speed boat has S
    constant velocity of (7i + 24j) ms-1.
    S B
    rs  600 j   7i  8 j t rB  120 j   7i  24 j t
    B
    b) Show that S and B will collide and find the time when
    this collision occurs and the position vector of the point of
    collision Observation Point
    i j
     7t   7t 600   8 t  120   24 t
    t  30
    So ships collide at t = 30, we then substitute into our position vectors to find value at
    time of collision
    check with the other vector…

    rs  600 j   7i  8 j t rB  120 j   7i  24 j t
    rs  600 j  30 7i  8 j rB  120 j  30 7i  24 j
    rs  210i  840 j rB  210i  840 j

    So the ships collide 30s after noon at 210i + 840j



    Question 11 Exercise 2D N

    At noon a ship S in 600m due north of an


    observation point O and a speedboat B is 120m due
    north of the same point. The ship S has constant
    velocity (7i + 8j) ms-1 and the speed boat has S
    constant velocity of (7i + 24j) ms-1.
    S B
    rs  600 j   7i  8 j t rB  120 j   7i  24 j t
    B
    In order to prevent a collision, 15s after noon S changes
    its velocity to (7i +30j) ms-1.
    Observation Point

    c) Find the distance between S and B 30s after noon

    Consider S at t = 15
    rs  600 j  15 7i  8 j
    rs  600 j  105i  120 j
    rs  105i  720 j
    Question 11 Exercise 2D N

    At noon a ship S in 600m due north of an


    observation point O and a speedboat B is 120m due S
    north of the same point. The ship S has constant
    velocity (7i + 8j) ms-1 and the speed boat has
    constant velocity of (7i + 24j) ms-1. B

    S B
    rs  600 j   7i  8 j t rB  120 j   7i  24 j t
    In order to prevent a collision, 15s after noon S changes
    its velocity to (7i +30j) ms-1.
    Observation Point

    c) Find the distance between S and B 30s after noon

    Consider S at t = 15
    rs  600 j  15 7i  8 j
    rs  600 j  105i  120 j
    rs  105i  720 j
    S changes course and travels (7i +30j) ms-1 for 15 seconds, so find S after this time
    rs  105i  720 j   7i  30 j t
    rs  105i  720 j  15 7i  30 j
    rs  210i  1170 j
    Question 11 Exercise 2D N

    At noon a ship S in 600m due north of an B


    observation point O and a speedboat B is 120m due
    north of the same point. The ship S has constant
    velocity (7i + 8j) ms-1 and the speed boat has
    constant velocity of (7i + 24j) ms-1.
    S B
    rs  600 j   7i  8 j t rB  120 j   7i  24 j t
    In order to prevent a collision, 15s after noon S changes
    its velocity to (7i +30j) ms-1.
    Observation Point

    c) Find the distance between S and B 30s after noon

    Now at t = 30
    rs  210i  1170 j rB  210i  840 j

    Same i value so we just consider the vertical

    Distance = 1170 – 840 = 330 m

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