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    Hi,
    I have this question.

    A stream provides a constant acceleration of 6 ms-2. A toy boat is pushed directly against the current and then released from a point 1.2m upstream from a small waterfall. Just before it reaches the waterfall, it is travelling at a speed of 5ms-1.

    a) Calculate the initial velocity of the boat
    b) Calculate the maximum distance upstream from the waterfall the boat reaches.

    I have done a)
    s = 1.2
    u = ?
    v = 5ms-1
    a = 6ms-2
    t

    v^2 = u^2 + 2as
    5^2 = u^2 + 2(6)(1.2)
    25 = u^2 + 14.4
    u^2 = 10.6
    u = -3.26 as the direction is negative. (The boat initially travels the other direction)

    But how do I do b)
    thanks.
    Note, the answer is 2.08m but it doesn't matter to me about the answer, but how to get there.
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    (Original post by nwmyname)
    Hi,
    I have this question.

    A stream provides a constant acceleration of 6 ms-2. A toy boat is pushed directly against the current and then released from a point 1.2m upstream from a small waterfall. Just before it reaches the waterfall, it is travelling at a speed of 5ms-1.

    a) Calculate the initial velocity of the boat
    b) Calculate the maximum distance upstream from the waterfall the boat reaches.

    I have done a)
    s = 1.2
    u = ?
    v = 5ms-1
    a = 6ms-2
    t

    v^2 = u^2 + 2as
    5^2 = u^2 + 2(6)(1.2)
    25 = u^2 + 14.4
    u^2 = 10.6
    u = -3.26 as the direction is negative. (The boat initially travels the other direction)

    But how do I do b)
    thanks.
    Note, the answer is 2.08m but it doesn't matter to me about the answer, but how to get there.
    at maximum distance upstream the velocity is zero, so one way would be to calculate the distance required to accelerate from 0 to 5 m/s at 6 m/s2
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    (Original post by nwmyname)
    Hi,
    I have this question.

    A stream provides a constant acceleration of 6 ms-2. A toy boat is pushed directly against the current and then released from a point 1.2m upstream from a small waterfall. Just before it reaches the waterfall, it is travelling at a speed of 5ms-1.

    a) Calculate the initial velocity of the boat
    b) Calculate the maximum distance upstream from the waterfall the boat reaches.

    I have done a)
    s = 1.2
    u = ?
    v = 5ms-1
    a = 6ms-2
    t

    v^2 = u^2 + 2as
    5^2 = u^2 + 2(6)(1.2)
    25 = u^2 + 14.4
    u^2 = 10.6
    u = -3.26 as the direction is negative. (The boat initially travels the other direction)

    But how do I do b)
    thanks.
    Note, the answer is 2.08m but it doesn't matter to me about the answer, but how to get there.
    B) so youve got initial velocity is 3.26, accelaration is -6. You know that at the maximum distance away from the waterfall v will be 0 (because it is changing direction at that moment). Then youre left with s as the variable. So you imput the numbers into v2=u2+2as. 0=10.6+(2x-6xs). 0=10-12s. S=10.6/12=0.88. You then add this to 1.2 since this is the distance it travels PAST the 1.2m and this gets you 2.08. A helpful note: its very common for questions to be asked where they dont give you a value for v or u when those values are 0. If you are asked about maximum distance you usually need to work out the distance when v is 0 as this is when a projectile will reach its highest point/maximum distance. Hope this helped
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    Joinedup CameronNicholson thank you very much, both of you! I got it now.
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    (Original post by nwmyname)
    Hi,
    I have this question.

    A stream provides a constant acceleration of 6 ms-2. A toy boat is pushed directly against the current and then released from a point 1.2m upstream from a small waterfall. Just before it reaches the waterfall, it is travelling at a speed of 5ms-1.

    a) Calculate the initial velocity of the boat
    b) Calculate the maximum distance upstream from the waterfall the boat reaches.

    I have done a)
    s = 1.2
    u = ?
    v = 5ms-1
    a = 6ms-2
    t

    v^2 = u^2 + 2as
    5^2 = u^2 + 2(6)(1.2)
    25 = u^2 + 14.4
    u^2 = 10.6
    u = -3.26 as the direction is negative. (The boat initially travels the other direction)

    But how do I do b)
    thanks.
    Note, the answer is 2.08m but it doesn't matter to me about the answer, but how to get there.
    At max height you know that the vertical final velocity is 0:

    Vertical:
    S- ?
    U- 5 (from the question)
    V- 0
    A- -6
    T- ?

    Horizonal:
    S- 1.2
    U- 3.26
    V- 3.26
    A- -6
    T- ?

    FIrst thing I did was work out the time taken using the vertical components:
    v-u/a= t --> 0-5/-6 = 0.83 s

    Now use this in another equation:
    s = 0.5(u+v)t --> 0.5(5)0.83 = 2.075 = 2.08 (3s.f)
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    (Original post by nwmyname)
    Hi,
    I have this question.

    A stream provides a constant acceleration of 6 ms-2. A toy boat is pushed directly against the current and then released from a point 1.2m upstream from a small waterfall. Just before it reaches the waterfall, it is travelling at a speed of 5ms-1.

    a) Calculate the initial velocity of the boat
    b) Calculate the maximum distance upstream from the waterfall the boat reaches.

    I have done a)
    s = 1.2
    u = ?
    v = 5ms-1
    a = 6ms-2
    t

    v^2 = u^2 + 2as
    5^2 = u^2 + 2(6)(1.2)
    25 = u^2 + 14.4
    u^2 = 10.6
    u = -3.26 as the direction is negative. (The boat initially travels the other direction)

    But how do I do b)
    thanks.
    Note, the answer is 2.08m but it doesn't matter to me about the answer, but how to get there.
    hi are you doing g481 do you have the 2015 ppaper and mark scheme?
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    (Original post by Mihael_Keehl)
    hi are you doing g481 do you have the 2015 ppaper and mark scheme?
    I can provide them... but I would need permission to do that from a higher power on this forum.
 
 
 
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