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    Stie's Avatar
    Stie Posts: 4, Reputation: 1
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    #1

    Jun 22, 2009, 09:49 AM
    Calculate the following:
    A body having a mass of 20kg accelerates uniformly in a straith line according to the law, v=10 - 2t, where v is the final velocity after t in seconds.
    Calculate the following:
    1.1.1 The initial velocity
    1.1.2 The acceleraton
    1.1.3 The initial momentum
    1.1.4 The kinetic energy after 4 seconds
    1.1.5 The distance travelled after 3 seconds
    There's more to the question but I don't know how to type in quadratics or any other sign for that matter.
    Please show all steps to above mentioned questions
    thanks
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #2

    Jun 22, 2009, 09:57 AM

    Initial velocity is when t=0

    Acceleration is the rate of change of velocity. You have the velocity when t=o, find the velocity when t=1, then, do:



    Momentum is given by mass x velocity

    Kinetic energy is . You have to find the velocity when time is 4 though.

    Distance travelled is found by the following formula: (s is displacement or distance, u is initial velocity, a is acceleration)



    Hope it helped! :)

    [Btw, we are not supposed to do your work, but help you do it]
    Stie's Avatar
    Stie Posts: 4, Reputation: 1
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    #3

    Jun 22, 2009, 10:13 AM
    Quote Originally Posted by Unknown008 View Post
    Initial velocity is when t=0

    Acceleration is the rate of change of velocity. You have the velocity when t=o, find the velocity when t=1, then, do:



    Momentum is given by mass x velocity

    Kinetic energy is . You have to find the velocity when time is 4 though.

    Distance travelled is found by the following formula: (s is displacement or distance, u is initial velocity, a is acceleration)



    Hope it helped! :)

    [Btw, we are not supposed to do your work, but help you do it]
    Thanks for helping. I know I have to do my own work. I am 24 and have to finish Mathematics N3/Matric and Engineering Science in less than a month. I did not take this subjects at school in my matric year so now I have to do 2 subjects in less than a month. The work that I am completing now is for my yearly marks and my x-sams start at the end of July 2009. These to worksheets that I have to complete is due tomorrow. Thanks anyway for u'r help. I don't know how i am going to complete it as you can see that I have absolutely know idea how to do it. So if there's any other way anyone else can help me please, i would appreciate it. :confused:
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #4

    Jun 22, 2009, 10:48 AM

    Ok, I'll do the first one, you'll see it's easy peasy!

    v=10-2t

    At the start, the time is 0, right? Then, put 0 in the place of t in your equation, you have;

    v = 10 - 2t
    v = 10 - 2(0)
    v = 10 - 0
    v = 10

    So, the initial velocity is 10 m/s.
    Stie's Avatar
    Stie Posts: 4, Reputation: 1
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    #5

    Jun 22, 2009, 11:12 AM
    Quote Originally Posted by Unknown008 View Post
    Ok, I'll do the first one, you'll see it's easy peasy!

    v=10-2t

    At the start, the time is 0, right? Then, put 0 in the place of t in your equation, you have;

    v = 10 - 2t
    v = 10 - 2(0)
    v = 10 - 0
    v = 10

    So, the initial velocity is 10 m/s.
    Thanks Jerry,

    I Still don't have a clue what to do. The last time I went to a Science class were about six years back. I entered my name and email address on the internet at Yahoo for a tutor. Science worksheet has 8 Questions. So hopefully somebody going to help me with my Science and Maths worksheets. I am going to be up all night as long as it takes, otherwise I'm just going to give up. If I can only complete these worksheets and understad them... than I have time until end July to learn for exams. Thanks
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #6

    Jun 22, 2009, 11:21 AM

    Ok, I understand your problem, I'll give you the explanations for the rest then.

    Acceleration is defined as the rate of change of velocity. Therefore, you need two velocities at two dissimilar times. You already found 10 m/s at t=0, and the easiest is to find when t=1, so, v = 10-2(1) = 8 m/s.

    Now,

    It's the same formula as for the gradient of a line. It's 'final' minus 'initial' over 'final' over 'initial'. Did that make sense to you?

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