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Help on motion and pressure needed

  1. Jun 21, 2005 #1
    1. A train accelerates from rest at a constant rate and reaches a velocity of 153 kmh in a distance of 2.29 km. Calculate the acceleration and time taken.
    If the breaks are then applied and produce a constant deceleration bringing the train to rest in a time of 140s, calculate the rate of deceleration

    2. A vertical cylindrical chemical process tank has a diameter of 3m. the tank has a capacity when full, of 51980 litres. if the tank is filled to 72% of its maximum capacity with a mixture of liquids with a combined density of 1063 kg/m3 determine the pressure at the base of the tank
  2. jcsd
  3. Jun 21, 2005 #2
    Welcome to PF. Their are people here much smarter and more learned that I, and in order to receive help it is expected that the poster show some work on his/her questions.

    I'll get you started on the first one...

    Use this nice kinematic equation to solve for "a".

    [tex]v_{f}^2 = v_{i}^2 + 2ad[/tex]

    You know everything but "a".

  4. Jun 21, 2005 #3
    some tip for the second q:
    72% of capacity should also mean 72% of the height. You have the diameter and capacity(volume) of the cylinder therefore you can find its height.
    Plug in pressure = hpg
  5. Jun 22, 2005 #4
    72% full it is 37,425.6 litres
    height is (51980 / 1000) / 7.1 = 7.3
    (i'm not sure if i am doing this right)
    i know that pressure= force/area and that density = mass/volume i just don't know how to apply it to the problem
  6. Jun 22, 2005 #5
    v2= u2+ 2ad
    v2-u2/2 = ad
    v2-u2/2d= a? (i don't think thats correct im not sure)
  7. Jun 23, 2005 #6
    pressure = force/area --- 1
    force in this case is the weight of the volume of liquid which is
    force(weight of liquid) = mass * g
    = density of liquid * volume * g
    = density of liquid * area * height of liquid * g --- 2
    sub 2 into 1 :
    pressure = (density of liquid * area * height of water * g) / area
    = density of liquid * height of liquid * g
    = p*h*g (simplified - this equation can be used for finding pressure in liquid at different height)
  8. Jun 23, 2005 #7
    no, it should be
    (v2-u2)/2 = ad
    (v2-u2)/2d= a?
  9. Jun 23, 2005 #8
    For qn1(a) To make this simpler, you can actually draw a velocity time graph to get u'r time knowing the area under the graph is the distance travelled.. After finding the time, u can use the formula [tex] a = v - u / t [/tex] to get acceleration.

    For the 2nd part to qn 1, u can find the deceleration by using the same formula in a

    For qn 2, first, juz know that [tex] P = pgh [/tex] .. Where the density is =1063[tex] Kg/m^3 [/tex], g =9.81[tex] m/s^2 [/tex] and the height 5294m
    Last edited: Jun 23, 2005
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