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Math Help - Velocity of a bullet

  1. #1
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    Velocity of a bullet

    Here is another one I am working on. I need help finding the linear equation, if I could, please. Once I have the equation, I can figure the rest out, I believe.

    Thanks in advance for your assistance.

    Velocity of a bullet.
    A gun is fired straight upward. The bullet leaves the gun at 100 feet per second (time t=0). After 2 seconds the velocity of the bullet is 36 feet per second. There is a linear equation that gives the velocity v in terms of the time t. Find the equation and find the velocity after 3 seconds.
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  2. #2
    Super Member Bacterius's Avatar
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    Alright. First, read the question and collect all information :

    - it is a linear equation, so it has the form y = mx + p (I believe it is useless to check if the line is parallel to the y-axis, that would be boring )
    - we are talking about velocity. Therefore, we are talking of feet per second.
    - the bullet leaves the gun at 100 feet per [1] second. Doesn't this remind us of something ? Yes, I guess that would be the slope of the line : if the bullet goes up 100 feet (y-axis) for 1 second (x-axis), then the slope will be rise/run = 100 / 1 = 100. Therefore, in the equation, m = 100.
    - After 2 seconds the velocity of the bullet is 36 feet per second. Hmm ... maybe this could actually be a point of the linear equation. Therefore, the point (2 ; 36) belongs to the equation.

    Good. We know that y = 100x + p, and that (2 ; 36) is a point of the equation. But we are still missing p to complete the equation. Let's try to substitute the values of the point that we have into the current equation to find p (since the point belongs to the equation) :

    36 = 100 * 2 + p
    36 = 200 + p
    36 - 200 = p
    p = -164

    We found p ! Now our equation is complete : y = 100x - 164

    Let's conclude : the velocity y of the bullet (in feet) at time x (in seconds) is given by the linear equation y = 100x - 164
    Note that I used x and y to denote the velocity and time. You can change following the question, therefore the equation is :

    v = 100t - 164

    The second question should be easy now
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    Thank you for your quick reply. I am calling it a night, but will look over your notes in the morning and apply the information you have supplied.

    I sincerely appreciate the help.
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  4. #4
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    Ok, I have worked though your figures. Thank you again.

    So, if v=100t-164 for the main equation, this is what I got for the velocity of the bullet after 3 seconds:

    v=100t-264

    I reached this by doing 36=100(3)+b
    (I am familiar with the linear equation of y=mx+b, so I followed that).

    So, 36=100(3)+b, to 36=300+b, to -264=b, or b=-264

    I substituted the letter v for y, and the letter t for x.
    Final answer: v=100t-264

    Would you concur?
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  5. #5
    Super Member Bacterius's Avatar
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    Nah, the equation is v = 100t - 164, and that's all. Now you want the velocity "v" for a time t = 3. Therefore, you substitute 3 into "t" in the equation :

    v = 100 * 2 - 164
    v = 200 - 164
    v = 36

    Therefore, at 3 seconds, the velocity of the bullet is 36 feet per second.
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  6. #6
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    Quote Originally Posted by Bacterius View Post
    Nah, the equation is v = 100t - 164, and that's all. Now you want the velocity "v" for a time t = 3. Therefore, you substitute 3 into "t" in the equation :

    v = 100 * 2 - 164
    v = 200 - 164
    v = 36

    Therefore, at 3 seconds, the velocity of the bullet is 36 feet per second.
    Wouldn't substituting 3 for 't' mean v=100*3-164?

    So v=300-164, to v=136?

    Therefore, at 3 seconds, the velocity of the bullet is 136 feet per second.

    As for clarifying the other part of the equation, thank you very much. I sincerely appreciate the help and clarification.
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  7. #7
    Super Member Bacterius's Avatar
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    Uhm yeah sorry it is 3 and not 2, typo.
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  8. #8
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    Quote Originally Posted by Bacterius View Post
    Uhm yeah sorry it is 3 and not 2, typo.
    Cool. No problem and thanks!
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