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    I have just started M1 and I am having a few problems on some of the questions. I have the heinemann M1 new edition text book so I will just state the questions I am having trouble with first. I will write them up in full later on but I am short on time ...

    Exercise 2D Q 11c:

    I get 300 and cant see why its wrong. My diagram shows that there is a straight line between them ???

    Exercise 3B Q 10:

    A stone is dropped from the top of a tower. One second later another stone is thrown vertically downwards from the same point with velocity 14 ms-1. If they hit the ground together find the height of the tower.

    Exercise 3C Q 8:

    Two cyclists A and B are travelling in the same direction along a straight road. B is travelling at a speed of 9 ms-1 and A is overtking B at a speed of 10ms-1. At the moment when they are level they see a traffic light turn red 108m. A cycles for Ts and then decelerates uniformly; B cycles for 6s and then decelerates uniformly. A and B stop at the same instant.
    a) sketch the speed-time graphs of the two cyclists on the same diagram.
    b) Calculate the time during which was decelerating
    c) Calculate the value of T

    I have manged to draw the speed time graph and it matches the answers. The problem is that I think that T can vary becuase the rate of decleration isnt stated. therefore T could be later on and A decelerates faster. Am I wrong in thinking this becuase I cant see a way to do this question!
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    I will try to help you with this one. Forgive me if I lack clarity, or make a mistake:
    Exercise 3B Q 10:

    A stone is dropped from the top of a tower. One second later another stone is thrown vertically downwards from the same point with velocity 14 ms-1. If they hit the ground together find the height of the tower.
    --------------------
    Taking g as 9.8ms^-2

    Dropped stone: s = ut + 1/2at^2 from the kinematics equations...
    s = 0 + 4.9t^2 (the 0 is due to it being dropped from rest).

    Thrown stone: s = ut + 1/2at^2
    s = 14t + 4.9t^2
    Because THIS stone is thrown one second later, let the value of t be (t - 1)
    So, s = 14t - 14 + 4.9(t-1)^2
    s = 14t - 14 + 4.9t^2 - 9.8t + 4.9
    s = 4.9t^2 + 4.2t - 9.1

    Combining the two equations...

    4.9t^2 + 4.2t - 9.1 = 14t + 4.9t^2

    9.8t = -9.1... UMMM NO?!

    That wasnt even the question, I did this yesterday! What the hell is my problem!?
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    (Original post by Womble548)
    Exercise 3B Q 10:

    A stone is dropped from the top of a tower. One second later another stone is thrown vertically downwards from the same point with velocity 14 ms-1. If they hit the ground together find the height of the tower.
    first stone:
    u=0, s=h, a=9.8
    s=ut + 0.5at^2

    h = 4.9t^2 -------(1)

    second stone:
    u=14, s=h, a=9.8
    T=(t-1), because it must travel in 1 less second than the other stone

    s=ut + 0.5at^2

    h = 14(t-1) + 4.9(t-1)^2
    h = 14t - 14 + 4.9t^2 - 9.8t + 4.8
    h = 4.9t^2 + 4.2t - 9.1

    substitute eq. (1):
    h = h + 4.2√(h/4.9) - 9.1
    √(h/4.9) = 9.1/4.2
    h = 4.9*(9.1/4.2)²
    h = 23.00m
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    (Original post by Womble548)
    .....

    Exercise 3C Q 8:

    Two cyclists A and B are travelling in the same direction along a straight road. B is travelling at a speed of 9 ms-1 and A is overtking B at a speed of 10ms-1. At the moment when they are level they see a traffic light turn red 108m. A cycles for Ts and then decelerates uniformly; B cycles for 6s and then decelerates uniformly. A and B stop at the same instant.
    a) sketch the speed-time graphs of the two cyclists on the same diagram.
    b) Calculate the time during which was decelerating
    c) Calculate the value of T

    I have manged to draw the speed time graph and it matches the answers. The problem is that I think that T can vary becuase the rate of decleration isnt stated. therefore T could be later on and A decelerates faster. Am I wrong in thinking this becuase I cant see a way to do this question!
    The rates of deceleration aren't stated, but they are fixed. Because both the distance and time are fixed this constrains when the deceleration should cut in and its value. We can work out that the overall time, from starting level to reaching the lights, is 18 sec. Try putting in different values of T then working out the deceleration required. Your eqns will be inconsistent.

    3C Q8

    Travel by B
    ========
    B travels for 6s at 9 m/s so travels for 54m
    This leaves 108 - 54 = 54 m travelled under a uniform deceleration of d_B m/s².
    B decelerates from 9 m/s to zero over a distance of 54m

    v² = u² - 2d_Bs
    0 = 81 - 2d_B*54
    d_B = 81/108 = 9/12 = 3/4
    d_B = 3/4 m/s²
    ==========

    Let t_B be the time travelled by B under uniform deceleration, then

    v = u + at
    0 = 9 - d_B.t_B
    9 = (3/4)t_B
    t_B = 12 s
    =======

    A and B both travel for the same amount of time, therefore

    T + t_A = 6 + 12
    T = 18 - t_A
    ========


    Travel by A
    ========
    A travels for Ts at 10 m/s so travels for 10T m
    This leaves (108 - 10T) m travelled under a uniform deceleration of d_A m/s².
    A decelerates from 10 m/s to zero over a distance of (108-10T) m

    v² = u² - 2d_As
    0 = 100 - 2d_A*(108-10T)
    d_A = 50/(108-10T)
    ==============

    Let t_A be the time travelled by A under uniform deceleration, then

    v = u + at
    0 = 10 - d_A.t_A
    10 = 50/(108-10T)*t_A
    108 - 10T = 5t_A

    substituting for T = 18 - t_A,

    108 - 10(18 - t_A) = 5t_A
    108 - 180 + 10t_A = 5t_A
    5t_A = 72
    t_A = 14.4 s
    ========

    T = 18 - t_A
    T = 18 - 14.4
    T = 3.6 s
    ======
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