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Homework Help: Truck problem (Kinematics)

  1. Nov 20, 2011 #1
    1. The problem statement, all variables and given/known data

    Farhana is late for class and she is ready to do anything to get to school on time. She knows a mattress truck passes her balcony exactly at 8:42 a.m, travelling at a constant 40km/h. If her balcony is 8.21m from the ground and the truck has a height of 3.47. How far must the truck be from the balcony for Farhana to land on it.

    Now i know, we have to find Δ d, and v1= 40km/h or 11.1111111111 m/s.

    2. Relevant equations

    Δd= v1Δt+ 1/2 aΔt^2

    3. The attempt at a solution

    Δd= 8.21m- 3.47m
    = 4.74m

    Δt= Δd/Δv
    = 4.74/ 11.111111111
    = 0.4266 s

    a= -9.8m/s^2

    v1= 11.111111111m/s

    Now it's asking how far the truck has to be, so it is asking for distance, and distance is speed over time. But i don't know how to find time. i tried but am i doing it right?
  2. jcsd
  3. Nov 20, 2011 #2
    So first step for you is to acknowledge that

    time for truck to get to the balcony = time that takes her to fall on top of the truck

    now for the girl to fall off can be derived by using the "five kinematic equation"
    we know the V1 = 0, we know delta d=4.74 we know acceleration is a=-9.8
    we can now solve for the time

    now once you have the time
    you know the equation distance = speed x time
    we know the speed of the truck and now we know time
    we can therefore get the distance
  4. Nov 20, 2011 #3
    I tried doing this and still got the wrong answer :S
  5. Nov 20, 2011 #4
    The answer is 11m

    and t= 0.569875104 or t= 1.697471834.

    if distance = speed * time

    and speed is 11.11m/s
    dist= 11.11 * 0.569875104
    dist= 6.331312405 (6.3m)

    and the other one is just wayy off.
  6. Nov 20, 2011 #5
    I got 11 m

    did you get time = 0.98 second?

    I think you got the quadratic equation wrong

    [tex]\Delta d = V_{i}\Delta t + ( 1/2)a(\Delta t)^2 [/tex]

    [tex] \Delta t = \frac{-V_{i} \pm \sqrt{ (V_{i})^2 - 4(1/2)a(-\Delta d)}}{a} [/tex]
  7. Nov 20, 2011 #6
    did you use v1= 0 or 11.11?
  8. Nov 20, 2011 #7
    tell me if i'm doing anything wrong,

    [tex]\Delta d = V_{i}\Delta t + ( 1/2)a(\Delta t)^2 [/tex]

    delta t = -11.11 +- sqrt (11.11)^2 -4(1/2)(-9.8)(-4.74)

    -11.11+- sqrt 30.5281

    ----------------- = 0.569875104

    -11.11 - 5.525223977
    --------------------- = 1.697471834

    am i right till here?
  9. Nov 20, 2011 #8
    [tex] V_{i} = 0 [/tex]
  10. Nov 20, 2011 #9
    To isolate t to one side

    [tex]\Delta d = V_{i}\Delta t + ( 1/2)a(\Delta t)^2 [/tex]

    [tex]0 = V_{i}\Delta t + ( 1/2)a(\Delta t)^2 -\Delta d [/tex]

    from here you use the quadratic equation which would be

    if [tex] y= ax^2 + bx +c [/tex]

    [tex] x= \frac{b \pm \sqrt {b^2 -4ac}}{2a} [/tex]

    in this case
    a= (1/2)(9.8)
    b= Vi
    c= -( delta d)
    x = delta t
  11. Nov 20, 2011 #10
    OH OK, thank you so much. I finally now get it!
  12. Nov 20, 2011 #11
    anytime :) I am glad you got it
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