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Linear Motion

  1. Sep 21, 2003 #1
    Hey
    Ok, so im having troule with yet another accelerated linear motion question...

    any help is desperately needed and very much appreciated :)

    So here it goes:

    (A) A particle is projected vertically upwards with velocityum/s and is at a height h after t1 and t2 seconds respectively. Prove that:
    t1.t2 = (2h)/g

    (B) A car accelerates uniformly from rest to a speed v m/s. It then continues at this constant speed for t seconds and then decelerates uniformly to rest.
    The average speed for the journey is (3v)/4

    (i) Draw a speed-time graph and hence, or other wise, prove that the time for the journey is 2t seconds.

    (ii) If the car driver had observed the speed limit of (1/2)v, find the least time the journey would have taken, assuming the same acceleration and deceleration as in (i).

    Ok, so for part a, ive been told that "setting those equal should make it easy to get the result". Basically what i took that up as meaning is " h=h " so "(-g/2)t1^2+ u t1 = (-g/2)t2^2+ u t2 " , which gives me two variables t1 and t2, equating to each other. Where am i going wrong here? Is there some thing about the times that im missing or maybe something else pretty obvious that im missing? When i try to "fix" my answer i get " (2h)/g = 8ut - 4t^2 ". Which is very very wrong...

    So that was part a, now onto part b:

    Ive drawn the graph, and made a very dodgy comp reproduction:

    Graph Pic

    All i can make out is that the area of space 1 is .5(t1)(v) = distance, area of space 2 is (t)(V) and area of space 3 is (.5)(t2)(v).
    So again ive got 3 variables (t1, t2 and V). V i can keep, since the average speed is given as being = (3v)/4 (i think this is right anyway). So now that ive done some very basic (very possible that its just basically wrong too...) stuff, can anybody give me a little more help ASAP?
    Thanks in advance
     
  2. jcsd
  3. Sep 21, 2003 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I suggested that, since h= (-g/2)t12+ u t1 and
    h= (-g/2)t22+ u t2 and you are asked about when the two have
    the same height, you try setting them equal:

    h=(-g/2)t12+ u t1(-g/2)t22+ u
    t2.

    Now, Combine those! (-g/2)(t12-t22)+
    u(t1-t2)= 0.


    Major hint!! FACTOR t12-t22.

    "All i can make out is that the area of space 1 is .5(t1)(v) = distance, area of space 2 is (t)(V) and
    area of space 3 is (.5)(t2)(v). So again ive got 3 variables (t1, t2 and V)."

    Actually, your graph is very good. You have a trapezoid with height v. The top "base" is t and
    the bottom base is t1+ t+ t2 so the area is (1/2)v(t1+ 2t+
    t2). The average speed is that area divided by the total time,
    t1+ t+ t2 that gives
    (1/2)v(t1+ 2t+ t2)/(t1+ t+ t2)= 3v/4 or
    (t1+ 2t+ t2)/(t1+ t+ t2)= 3/2.
    (t1+ t2+ 2t)= (3/2)(t1+t2) + (3/2)t so
    (1/2)(t1+ t2)= -(1/2) t. That is, t1+ t2= t. The total
    time the vehicle is moving is (t1+ t2)+ t= 2t !!!!
     
  4. Sep 21, 2003 #3
    Part A:

    Ok, so im trying to prove that t1.t2 = (2h)/g , and ive got to try and get that from "h=(-g/2)t12+ u t1(-g/2)t22+ ut2" - even with your "major hint" (?!?) im a little lost still. My brain has gone into serious meltdown over the summer vacation.... Ive got a very very vague idea of how it *might* come out, but i need another "major" hint at least.

    Part B:
    (i)
    We dont actually have trapezoids on our curriculum, so i was trying to take the area in 3 segments, as you can see from my diagram, which added *slightly* to the confusion.
    Towards the end of the proof you say
    Im a little confused as to where the minus sign went after you multiplied accross by two. Is it just because a minus answer wouldnt make any sense or is there something else im not getting?

    (ii)
    If the car driver had observed the speed limit of (1/2)v, find the least time the journey would have taken, assuming the same acceleration and deceleration as in (i).

    Regarding this one, im a little unsure as how to start off, i know the the maximum speed is now (1/2)v, instead of v in the last question... Thats about as much as i know as being true, regading the rest, im pretty lost, help! please!

    Thanks again for the help and send some more my way asap!
     
    Last edited: Sep 22, 2003
  5. Sep 22, 2003 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    (A) A particle is projected vertically upwards with velocityum/s and is at a height h after t1 and t2 seconds respectively. Prove that:
    t1.t2 = (2h)/g

    t1 and t2 are both solutions to the quadratic equation

    (-g/2)t^2+ vt= h which is the same as (g/2)t^2- vt+ h= 0
    which is again the same as t^2- (2v/g)t+ 2h/g= 0

    Saying that t1 and t2 satisfy t^2- (2v/g)t+ 2h/g= 0 means that

    (t-t1)(t-t2)= t^2- (2v/g)t+ 2h/g.

    Multiply that out and see what happens.
     
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