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Please help

Scheduled Pinned Locked Moved Algorithms
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  • A Offline
    A Offline
    A
    wrote on last edited by
    #1

    I have four equations: s = e^0.245t (t = time(sec) over 10 seconds) s = distance(metres) s = 4.2ln(t + 1.2) s = 0.2t(t-5.1) s = 0.2t(t-5.1)(t-9.1) and for each equation I need to find the displacement,velocity, acceleration, start postition and how long each takes to reach 10 metres away from it's starting point. I know that: Velocity is: ds/dt Acceleration is :dv/dt I have no problem finding the Velocity and acceleration: 1. s = e^0.245t (Is this the displacement?) Velocity = .245 e^.245t Acceleration = (.245)^2 e^.245t 2. s = 4.2ln(t + 1.2) (Is this the displacement?) Velocity = 4.2* (1/t +1.2) Acceleration = -4.2 (t+1.2)^-2 3. s = 0.2t(t-5.1) (Is this the displacement?) Velocity= 0.2t^2 - 1.0t (is this right?) Acceleration = 0.4t-1.02 4. s = 0.2t(t-5.1)(t-9.1) (Is this the displacement?) =-(0.2t^2 - 1.02t)(t-9.1) =0.2t^3 - 1.82t^2 - 1.02t^2 + 9.282t =0.2t^3 - 2.84t^2 +9.282t Is this the velocity? Velocity =.6t^2 - 5.68t + 9.282 Is this the acceleration? Acceleration = 1.2t-5.68 Now here's the question, how do I find the starting position that each equation starts at, and how do i find how long it took the equation to finish. Thanks for any help!:) If you need me to explain something better please ask. Ps: Sorry for the long question.

    My blog:[^]

    C N 2 Replies Last reply
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    • A A

      I have four equations: s = e^0.245t (t = time(sec) over 10 seconds) s = distance(metres) s = 4.2ln(t + 1.2) s = 0.2t(t-5.1) s = 0.2t(t-5.1)(t-9.1) and for each equation I need to find the displacement,velocity, acceleration, start postition and how long each takes to reach 10 metres away from it's starting point. I know that: Velocity is: ds/dt Acceleration is :dv/dt I have no problem finding the Velocity and acceleration: 1. s = e^0.245t (Is this the displacement?) Velocity = .245 e^.245t Acceleration = (.245)^2 e^.245t 2. s = 4.2ln(t + 1.2) (Is this the displacement?) Velocity = 4.2* (1/t +1.2) Acceleration = -4.2 (t+1.2)^-2 3. s = 0.2t(t-5.1) (Is this the displacement?) Velocity= 0.2t^2 - 1.0t (is this right?) Acceleration = 0.4t-1.02 4. s = 0.2t(t-5.1)(t-9.1) (Is this the displacement?) =-(0.2t^2 - 1.02t)(t-9.1) =0.2t^3 - 1.82t^2 - 1.02t^2 + 9.282t =0.2t^3 - 2.84t^2 +9.282t Is this the velocity? Velocity =.6t^2 - 5.68t + 9.282 Is this the acceleration? Acceleration = 1.2t-5.68 Now here's the question, how do I find the starting position that each equation starts at, and how do i find how long it took the equation to finish. Thanks for any help!:) If you need me to explain something better please ask. Ps: Sorry for the long question.

      My blog:[^]

      C Offline
      C Offline
      cp9876
      wrote on last edited by
      #2

      so you are given s=f(t) and you are asked to find the starting position and the time taken to travel 10m. I think you should review your lecture notes one more time and figure it out for yourself. Look at some example that the lecturer did and see where the starting position is, and how to work out how far something travelled.


      Peter "Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."

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      • A A

        I have four equations: s = e^0.245t (t = time(sec) over 10 seconds) s = distance(metres) s = 4.2ln(t + 1.2) s = 0.2t(t-5.1) s = 0.2t(t-5.1)(t-9.1) and for each equation I need to find the displacement,velocity, acceleration, start postition and how long each takes to reach 10 metres away from it's starting point. I know that: Velocity is: ds/dt Acceleration is :dv/dt I have no problem finding the Velocity and acceleration: 1. s = e^0.245t (Is this the displacement?) Velocity = .245 e^.245t Acceleration = (.245)^2 e^.245t 2. s = 4.2ln(t + 1.2) (Is this the displacement?) Velocity = 4.2* (1/t +1.2) Acceleration = -4.2 (t+1.2)^-2 3. s = 0.2t(t-5.1) (Is this the displacement?) Velocity= 0.2t^2 - 1.0t (is this right?) Acceleration = 0.4t-1.02 4. s = 0.2t(t-5.1)(t-9.1) (Is this the displacement?) =-(0.2t^2 - 1.02t)(t-9.1) =0.2t^3 - 1.82t^2 - 1.02t^2 + 9.282t =0.2t^3 - 2.84t^2 +9.282t Is this the velocity? Velocity =.6t^2 - 5.68t + 9.282 Is this the acceleration? Acceleration = 1.2t-5.68 Now here's the question, how do I find the starting position that each equation starts at, and how do i find how long it took the equation to finish. Thanks for any help!:) If you need me to explain something better please ask. Ps: Sorry for the long question.

        My blog:[^]

        N Offline
        N Offline
        Nathan Addy
        wrote on last edited by
        #3

        Suppose you have a function describing position of a particle in meters at a time: f(t) = e^(3t^2) Then, as you say, the first and second derivatives will be the position: v(t) = 6 t e ^(3t^2) and a(t) = 6 e ^ (3t^2) + 36 t ^2 e ^ (3t^2). To see how displacement works, plug in some values: f(0) = 1m f(.5) = 6.351m f(1) = 120.513m So in the first .5 seconds, the particle is at position 6.351m, but since it started at position 1m, it has only traveled a distance of 5.351m. I don't want to give it away, so I won't say more, but that should give you an idea of how to figure out the problem.

        A 1 Reply Last reply
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        • N Nathan Addy

          Suppose you have a function describing position of a particle in meters at a time: f(t) = e^(3t^2) Then, as you say, the first and second derivatives will be the position: v(t) = 6 t e ^(3t^2) and a(t) = 6 e ^ (3t^2) + 36 t ^2 e ^ (3t^2). To see how displacement works, plug in some values: f(0) = 1m f(.5) = 6.351m f(1) = 120.513m So in the first .5 seconds, the particle is at position 6.351m, but since it started at position 1m, it has only traveled a distance of 5.351m. I don't want to give it away, so I won't say more, but that should give you an idea of how to figure out the problem.

          A Offline
          A Offline
          A
          wrote on last edited by
          #4

          Thanks for the help.

          My blog:[^]

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