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Motion Of Bodies Coursework

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Ok I have been given an assignment and asked to answer questions about a lorry loaded and unloaded. the Text is as follows:
    An unloaded lorry of mass M kg accelerates from rest and reaches a velocity of 30mph in 10 seconds. It then travels at constant velocity for half a minute before the driver applies the brakes and brings the vehicle to rest over a distance of D m.
    The lorry is then loaded with 800 kg of building materials that need to be transported over a total distance of 0.6 km. After loading, the lorry accelerates over a distance of 80m to attain a velocity of 30mph again. The vehicle subsequently maintains constant velocity before decelerating to rest at a rate of R m/s2.

    M=2300KG D = 70m R = 1.5m/s^2


    2. Relevant equations
    The questions exactly
    Find for the unloaded lorry:

    i) the force required to accelerate the vehicle, and the retarding force necessary to bring it to rest;
    ii) the total time and the total distance travelled during this period.

    Find for the loaded lorry :

    i) the force required to accelerate the vehicle and the retarding force necessary to bring it
    to rest;

    ii) the total time taken during this period and the distance travelled at constant velocity.

    c) Calculate the % change in the momentum of the lorry from loaded to unloaded, whilst travelling at the specified constant velocities.

    d) Draw the ‘velocity-time’ graph of the motion of the lorry in its unloaded and loaded states, indicating clearly salient points and values.

    3. The attempt at a solution
    I tried using V^2 = U^2 +2as , V=S/t and general thinking but im 100% sure that im incorect. i got answers like 0.65 m/s^2 which seemed possible but it is half the acceleration than when the lorry was unloaded, i think half is too little if only 800KG was added. i also got stupid answers like 15.67 m/s^2

    can some one help me
     
  2. jcsd
  3. Feb 21, 2010 #2

    kuruman

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    Let us be systematic and do one thing at a time. Show your work for answering these questions for the unloaded lorry: Find

    1. the force required to accelerate the vehicle
    2. the retarding force necessary to bring it to rest
    3. the total time travelled during this period.
    4. the total distance travelled during this period.
     
  4. Feb 22, 2010 #3
    For the force Required for the unloaded lorry:

    A=V/T
    A= ((30 x 1.6093)x1000) / 3600
    A= 13.41 / 10
    A= 1.341 m/s2

    F=ma
    F= 2300 x 1.341
    F= 3084.3 N

    The retarding force

    A= (v2-u2) / 2s
    A= (0 - 13.41^2) / 2(70)
    A= 179.83 / 140
    A= - 1.28m/s2

    F=ma
    F= 2300 x -1.28
    F= -2944N

    Time :
    the first strech : 10s
    Constant streach : 30s
    Deceleration : A=V/T , T = V/A , T = 48.27/-1.28 , T = 37.71
    total : 77.71s

    distance : 1st = 13.41m (accelerated for 10s at 1.341 m/s^2)
    2nd = constant for 30s at 48.27kph so it traveled 402.25m
    3rd = 70m (stated in the question)
    total 485.60m

    Im not sure but i think this would all be right, but its B that gets me :(
     
  5. Feb 22, 2010 #4

    kuruman

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    Everything looks OK to this point. When the lorry accelerates from rest, is the distance traveled acceleration times time? What is the correct equation?
     
  6. Feb 22, 2010 #5
    thats the problem i face. it says in the background info that:
    The lorry is then loaded with 800 kg of building materials that need to be transported over a total distance of 0.6 km. After loading, the lorry accelerates over a distance of 80m to attain a velocity of 30mph again.

    that is all the informaition i am given, i used the equation:
    V^2=U^2-2as
    and derrived the equation for a, using the 80m as s.. but this gave me 14.56m/s^2 for the acceleration and i doubt this is it. am i right in saying the change in velocity was 48.27 kmph ? is there another equation i can use for this?
     
  7. Feb 22, 2010 #6
    I think i may have figured out B) :)
    The force required to accelerate the loaded lorry

    A= (v2-u2) / 2s
    A= (13.41^2 - 0) / 2(80)
    A= 179.83 / 160
    A= 1.12m/s2

    F=ma
    F= 3100 x 1.12
    F= 3472N

    The retardant force. (the deceleration was given in the question)

    F=ma
    F= 3100 x -1.5
    F= -4650N

    now the total time taken and distance. the distance had to worked out first to achieve the total time
    the first distance was given as 80m
    then the third had to be worked out:
    s = V^2 - U^2 /2a
    s = 0^2 - 48.27^2 / -3
    s = 59.94m

    adding all the distances i got 139.94 and take it away from the total distance of 0.6km i got 460.06m.

    1st leg of the journey
    t = V/A
    t = 48.27/1.12
    t = 43.1s

    2nd leg
    460.06 / 13.41m/s
    = 34s

    3rd leg
    t=v/a
    t=48.27/-1.5
    t=32.18s

    i got the total time to be 109.28s.

    am i correct?
     
  8. Feb 22, 2010 #7

    kuruman

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    So far so good for part B.

    How do you figure 48.27? How fast is the loaded lorry moving when it starts braking?

    Also note that you haven't fixed the initial distance when the lorry starts accelerating from rest in part A.
     
  9. Feb 22, 2010 #8
    the lorry is traveling at a constant velocity of 30MPH which in turn works out to be 48.27KPH.

    which part A? can you quote it :) thanks
     
  10. Feb 22, 2010 #9

    kuruman

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    You need to convert to m/s for the answer to come out in meters. The denominator is in m/s2.
    I already did. Look at posting #4.
     
  11. Feb 23, 2010 #10
    to change that in to meters would make it 13.41 m/s would that be correct?
     
  12. Feb 23, 2010 #11

    kuruman

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    Where did this come from in posting #3?
    Did you write it or did you copy it from somewhere?
     
  13. Feb 23, 2010 #12
    i wrote this... it was to change the 30mph to kph then to m/s then to acceleration
     
  14. Feb 23, 2010 #13

    kuruman

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    Then you know the answer to what you just asked.
     
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