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Acc Linear Motion

  1. Sep 29, 2003 #1
    Hey
    Just need a little help with *another* accelerated linear motion question, any help is appreciated:

    (A) A particle starts from rest at a point p and accelerates at 2m/s^2 until it reaches a speed v m/s.
    It travels at this speed for one minute before delerating at 1m/s^2 to rest at q. The total time for the journey is 2 minutes.
    (i) calculate the distance pq
    (ii)If a second particle starts from p at time t=0 and moves along pq with speed (2t + 50) m/s, find the time taken to reach q.

    Part one ive gotten out - pq = 3600m
    Part two, im not really too sure where to start.

    (B) A particle P is projected vertically upwards with an initial velocity u and two seconds later a second particle Q is projected vertically upwards from the same point with initial velocity 1.5u. Calculate, in terms of u, how long W is in motion before it collides with p, and prove that u > 9.8.

    Ok, so ive made a bit of a start on this one, whether or not its right is what im trying to find out, and what i go about doing next:
    Eqn1 : S=ut+(1/2)at^2
    Eqn2 : S=ut-(1/2)gt^2

    S=ut-(1/2)gt^2
    s=(1.5u)(t+2) - (1/2)g(t+2)^2
    s=1.5ut+3u-(1/2)g(t^2+4t+4)
    s=1.5ut+3u-(1/2)gt^2-2g-2gt

    Collision occurs when s=s
    ut-(1/2)gt^2 = 1.5ut+3u-(1/2)gt^2-2g-2gt
    ut=1.5ut+3u-2gt-2g
    2ut+3ut+6u-4gt-4g
    -ut+4gt=6u-4g
    t(4g-u)=6u-4g

    t= (6u-4g)/(4g-u)

    So thats all that ive got so far, any help very appreciated
    Thanks again
     
  2. jcsd
  3. Sep 30, 2003 #2

    HallsofIvy

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    Science Advisor

    For A1, I get your answer.

    Let t1 be the time the object accelerates at 2 m/s^2. At the end of that time it will have a speed of 2t1 and have gone a distance
    d1= (1/2)(2)t1^2= t1^2.

    It now goes for 60 s at that speed. It will have gone a distance
    d2= (2t1)(60)= 120 t1.

    Let t2 be the time it decelerates at 1 m/s^2. Since it's initial speed was 2t1, at the end of that time we have 2t1- (1)t2= 0 or
    t2= 2t1 (since the deceleration is half the acceleration, it takes twice as long). During that time it will have gone a distance
    d3= (2t1)t2- (1/2)(1)t2^2= 2t1(2t1)-(1/2)(2t1)^2= 4t1^2- 2t1^2
    = 2t1^2.

    We are told that the total time: t1+ 60+ t2= 120 so
    t1+ t2= t1+ 2t1= 3t1= 60. t1= 20 s and t2= 40 s.

    d1= (20)^2 = 400 m.
    d2= 120(20)= 2400 m.
    d3= 2(20)^2= 800 m.
    total distance=3600 m.

    A2 Ought to be easy. You now know that the distance to be traveled by the second object is 3600 m and you are told that the speed at time t is 2t+ 50 m/s. If you have taken calculus, distance is the integral of speed with respect to time:
    &int (t=0 to t1)(2t+ 50)dt= (t^2+ 50t)evaluated from t1 to 0:
    t1^2+ 50t1 and that must be 3600 m. Solve t1^2+ 50t1= 3600.

    If you have not taken calculus you could still argue: v(t)= 2t+ 50 means that the object had an initial speed of v0= 50 m/s and then accelerated at a= 2 m/s^2. We have the formula (derived by integrating!) d= (1/2)a t^2+ v0 t which, here, gives
    3600= t^2+ 50t exactly as before.

    In B you have a couple of minor errors.
    Since the second object is projected upward 2 seconds after the first, and t is measured from the moment the first one was projected upward, its "time after starting" is t-2, not t+2 as you have.
    Your equations should be S=ut-(1/2)gt^2
    s=(1.5u)(t-2) - (1/2)g(t-2)^2
    so that s= 1.5ut- 3u- (1/2)gt^2+ 2gt- 2.
    You are correct that they will collide when S= s.

    ut- (1/2)gt^2= 1.5ut- 3u- (1/2)gt^2+ 2gt- 2 and the t^2 terms cancel
    ut= 1.5ut- 3u+ 2gt- 2 or (.5+ 2g)ut= 3u+ 2 so t= (3u+2)/(.5+2g)u
    or (multiply numerator and denominator by 2) t= (6u+4)/((1+4g)u)
     
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