Turn on thread page Beta

M2 watch

Announcements
    Offline

    0
    ReputationRep:
    if u use t=4
    then v=8i+4j
    which is in the same ratio thus parallel to 2i+j
    • Thread Starter
    Offline

    15
    ReputationRep:
    As youve probably guessed Im struggling with this
    so:

    A particle X, moving along a straight line with constant speed 4m/s, passes through a fixed point O. Two seconds later another particle Y, moving along the same straight line and in the same direction, passes through O with speed 6m/s.
    Given that Y is subject to a cosntant deceleration of magnitude 2m/s/s

    a) state the velocity and displacement of each particle t seconds after Y passed through O

    b) find the shorest distance between the particles after they have both passed through O

    c) find the value of t when the distance between the particle has increased to 23m

    I struggled through a, didnt really understand it though, and dont know what to do for b/c. Can anyone go through it for me please
    • Thread Starter
    Offline

    15
    ReputationRep:
    Bump
    Offline

    2
    ReputationRep:
    (Original post by imasillynarb)
    As youve probably guessed Im struggling with this
    so:

    A particle X, moving along a straight line with constant speed 4m/s, passes through a fixed point O. Two seconds later another particle Y, moving along the same straight line and in the same direction, passes through O with speed 6m/s.
    Given that Y is subject to a cosntant deceleration of magnitude 2m/s/s

    a) state the velocity and displacement of each particle t seconds after Y passed through O
    i'm not too sure if i've done this right, but here's a) :

    vX = 4m/s
    vY = (6-2t) m/s

    when t=2, rX = 8m , rY = 0m
    so at time t seconds, rX = (8 + 4t) m
    rY = (6-2t)t = (6t - 2t^2)m
    Offline

    2
    ReputationRep:
    (Original post by imasillynarb)

    b) find the shorest distance between the particles after they have both passed through O
    i think you would just do (rX-rY)

    D = 8+4t - (6t - 2t^2)
    D = 2t^2 - 2t + 8

    to find min. differentiate: dD/dt = 4t - 2 = 0

    Therefore t=0.5 (to prove it's a min, differentitate again to get 4 => min)

    sub t=0.5, D= 7.5m
    Offline

    2
    ReputationRep:
    for last one, make D = 23, and get a quadratic to solve t. i get t= [1+rt(31)]/2
 
 
 

University open days

  • University of Warwick
    Undergraduate Open Days Undergraduate
    Sat, 20 Oct '18
  • University of Sheffield
    Undergraduate Open Days Undergraduate
    Sat, 20 Oct '18
  • Edge Hill University
    Faculty of Health and Social Care Undergraduate
    Sat, 20 Oct '18
Poll
Who is most responsible for your success at university

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.