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Math Help - Velocity, Acceleration, Displacement, Time

  1. #1
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    Velocity, Acceleration, Displacement, Time

    Greetings.

    I have missed many of my physics classes due to a formidable cold, and thus am in need of guidance on the following questions:

    1) Bill is playing in the backyard when his mother asks him to run to the letterbox and collect the mail. Figure 10.16 shows the velocity-time graph for his run to the letterbox.
    a. What is Bill's displacement: i. after 8.0 s? ii. between t = 10 s and t = 12 s.
    b. How far is the letterbox from where Bill was playing?
    c. What is the acceleration at: i. t = 6.0 s? ii. t = 14.0 s?



    2) An object starts from rest and travels in a straight line for 5.0 s with a uniform acceleration of 4.0 ms-2. It continues to move with the uniform velocity acquired for a further 12 s and it is finally brought to rest with a
    deceleration of 2.5 ms-1.
    a. Calculate the distance travelled and the time taken.
    b. Draw a velocity-time graph and show how the result for part a could also be obtained from the graph. (If you could simply tell me how this would work, that would be great.)

    Thanks in advance.
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    Re: Velocity, Acceleration, Displacement, Time

    The graph you provided appears to be an inverted parabola. It is velocity vs time. That meas the motion is not uniformly accelerated. You cannot use the standard kinematics formulae. The maximum time is about 19 s (the image is fuzzy at the point). Now you need to find the equation of the velocity v(t).

    since v(0) = 0, and v(19) = 0, you could model the velocity as v(t) = -kt(t-19) (Why negative sign in front of k?)
    Use the given data and find the equation.

    The displacement from point A to point B is going to be the inegral from A to B of v(t) = dx/dt.
    Work out that part and let me know if you need more help.
    Thanks from topsquark and Fratricide
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    Re: Velocity, Acceleration, Displacement, Time

    We haven't covered calculus yet, so I have no idea how to find the 'inegral from A to B of v(t) = dx/dt.'

    Also, the graph ends at t = 20 s and doesn't appear to be a perfect parabola, making me unsure as to whether your equation would work.
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    Re: Velocity, Acceleration, Displacement, Time

    Quote Originally Posted by Fratricide View Post
    We haven't covered calculus yet, so I have no idea how to find the 'inegral from A to B of v(t) = dx/dt.'

    Also, the graph ends at t = 20 s and doesn't appear to be a perfect parabola, making me unsure as to whether your equation would work.
    Then procede that way: draw the grid as shown in attached image on your graph; the time axis on your image is not horizontal. The intersection point of a vertical line at time t, t=12 s, intersets with the graph at a point that corresponds to a velocity v(t). Knowing that the displacement fromula at t is v(t)*t, the displacement between two times on the graph is the area under the graph v(t) from t1 to t2 with t2 > t1 becomes the area of the trapezoid. You will have to add the areas of all the trapezoids between t1 and t2 to obtain the displacement from t1 to t2.

    Let's see how far you could go.

    Velocity, Acceleration, Displacement, Time-untitled2.gif
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    Re: Velocity, Acceleration, Displacement, Time

    Okay, so I drew the grid (albeit somewhat messily):

    Spoiler:

    If I understand correctly, when t = 8 s, v(t) = 7.5 ms-1 (roughly). Therefore, displacement = v(t)*t = 7.5 x 8 = 60 m. The book reads "18 m (approx)".

    As for the second displacement question, I found the area of the trapezoid between t = 10 s and t = 12 s by doing ((8+7.5)/2)*2 = 15.5 m. The book reads "14 m (approx)".

    ---------------------------------

    EDIT: I had another look at the second question from my original post and found the correct answer. Well, half of it -- the answer for the total time taken wasn't in my textbook.

    First, I found the velocity given the acceleration, which was 20 ms-1. Next, I used the equation v2 = u2 + 2ad to find d. However, in order to get the right answer, I had to use 4 as the value for a, which was the acceleration at the start of the object's movement. Why, then, is the deceleration not included somewhere in the equation? One would assume that you would need to include it in order to find the total distance travelled, as the distance would decrease or increase depending on how quickly the objected stopped moving.

    Also, I used to the equation t = (v-u)/a to calculate t during the deceleration stage. I got an answer of 8 and added it to 17 to get a total time of 25 s for the object's movement. Is this correct?
    Last edited by Fratricide; October 3rd 2013 at 07:00 PM.
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    Re: Velocity, Acceleration, Displacement, Time

    Quote Originally Posted by Fratricide View Post
    Okay, so I drew the grid (albeit somewhat messily):

    Spoiler:

    If I understand correctly, when t = 8 s, v(t) = 7.5 ms-1 (roughly). Therefore, displacement = v(t)*t = 7.5 x 8 = 60 m. The book reads "18 m (approx)".

    As for the second displacement question, I found the area of the trapezoid between t = 10 s and t = 12 s by doing ((8+7.5)/2)*2 = 15.5 m. The book reads "14 m (approx)".

    ---------------------------------

    EDIT: I had another look at the second question from my original post and found the correct answer. Well, half of it -- the answer for the total time taken wasn't in my textbook.

    First, I found the velocity given the acceleration, which was 20 ms-1. Next, I used the equation v2 = u2 + 2ad to find d. However, in order to get the right answer, I had to use 4 as the value for a, which was the acceleration at the start of the object's movement. Why, then, is the deceleration not included somewhere in the equation? One would assume that you would need to include it in order to find the total distance travelled, as the distance would decrease or increase depending on how quickly the objected stopped moving.

    Also, I used to the equation t = (v-u)/a to calculate t during the deceleration stage. I got an answer of 8 and added it to 17 to get a total time of 25 s for the object's movement. Is this correct?

    v(t) is the instantaneous velocity, i.e. it is a function, not just one number on the graph. If you do not know the function v(t) the answer from v(t)*t is incorrect.

    Draw a triangle base = 8 and height = 7.5, those are data from your graph. The displacement from 0 to 8 would be the area of the triangle which is 30. Because the graph is above the straight line from 0 to 7.5, the area under the graph should be a little more than 30 m. I do not know where your book got 18 m. Either you provided incorrect question, or the 18 m of the book is incorrect.

    In the second problem you will need to work out each portion of the motion separately. First portion of the motion is 5 s with constant acceleration 4 m/s^2. At that point his velocity is v(5). At the 5th second a new portion of the motion begins, a uniform rectilinear motion with constant velocity exactly v(5) for another 12 s. At the end of the 12th of rectilinear motion its velocity is still v(5). This is the initial velocity of the third portion of the motion. There the acceleration is negative (you call it deceleration). You donít have the time for this portion of the motion but you have the acceleration and the initial velocity v(5); the final velocity 0 (at rest). You find the time, then calculate the distance traveled in the third portion of the motion. Add the distance from the three portions of the motion, and the times from the three portions of the motions to complete the answer to your problem
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    Re: Velocity, Acceleration, Displacement, Time

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