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Math Help - Mechanics Speed Question

  1. #1
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    Exclamation Mechanics Speed Question

    A balloon moves vertically upwards in a straight line so that at time t seconds its acceleration, a metre per second is
    a=3t/200

    Initally the balloon has a speeed of 0.5m/s
    i)find the speed of the balloon when t=30sec
    ii)Find how far the balloon travels in the first 30sec.

    Any help much appreciated.
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  2. #2
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    hi there,

    we have the following data

    <br />
Acceleration,a = \frac{3t}{200}<br />
    <br />
 InitialSpeed, u= 0.5<br />
    <br />
TimeTaken, t = 30<br />

    to find the final velocity we will use the formula
    <br />
v = u+at<br />

    substituting values we have the final velocity can be obtained


    to find the distance traveled for this first 30 seconds use this formula with the given information:
    <br />
s = ut+\frac{1}{2}at^2<br />

    <br />
Acceleration,a = \frac{3t}{200}<br />
    <br />
 InitialSpeed, u= 0.5<br />
    <br />
TimeTaken, t = 30<br />

    again by substituting the given value you should be able to get the required distance traveled ^-^ good luck.
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  3. #3
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    yes i have tried using those equationns previously but got answers of 13 etc. the answer i know is infact 7.25.help plz!!
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  4. #4
    Member Danshader's Avatar
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    ou ok i understand why it is not this method, the previous method was for constant acceleration, however in this case it is for varying acceleration:

    the right method to use is to consider:
    <br />
\frac{d}{dt}v = a<br />

    <br />
\frac{d}{dt}v = \frac{3t}{200}<br />

    <br />
\int  \frac{d}{dt}v = \int \frac{3t}{200}dt<br />

    <br />
v =  \frac{3t^2}{400} + c \longrightarrow(1)<br />

    when t = 0, u = 0.5, from (1);
    <br />
  u =  \frac{3(0)^2}{400} + c<br />
    <br />
c = 0.5<br />

    so the equation to determine the speed of the balloon after 30 seconds is:
    <br />
  v =  \frac{3t^2}{400} + 0.5<br />

    <br />
  v =  \frac{3(30)^2}{400} + 0.5<br />

    this will give you the answer of 7.25 ms^(-1)

    for the second part we can also observe just like in the case of finding the velocity:

    <br />
\frac{d}{dt}S = v<br />

    <br />
\int \frac{d}{dt}S =\int vdt<br />

    <br />
\int \frac{d}{dt}S =\int  \frac{3t^2}{400} + 0.5dt<br />

    <br />
 S = \frac{3t^3}{1200} + 0.5t + k \longrightarrow(2)<br />

    at t = 0, S = 0, therefore from equation (2):
    <br />
 S = \frac{3t^3}{1200} + 0.5t + k <br />

    <br />
  0 = \frac{3(0)^3}{1200} + 0.5(0) + k<br />
    <br />
 k = 0<br />

    therefore the general equation of the displacement, S is:

    <br />
  S = \frac{3t^3}{1200} + 0.5t <br />
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