{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Review Sheet _2_1 - x-direction and in the y-direction...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Rev. 2 REVIEW SHEET #2: Linear Motion Definitions of kinematic quantities: Position vector relative to origin ˆ ˆ ˆ r xi yj zk = + + r Displacement (vector) 2 1 ˆ ˆ ˆ r r r xi y j z k = - = ∆ + ∆ + ∆ r r r Velocity (vector) 2 1 2 1 ˆ ˆ ˆ ave r r r x y z v i j k t t t t t t - = = = + + - r r r r ˆ ˆ ˆ ˆ ˆ ˆ inst x y z dr dx dy dz v i j k v i v j v k dt dt dt dt = = + + = + + r r Acceleration (vector) 2 1 2 1 ˆ ˆ ˆ y x z ave v v v v v v a i j k t t t t t t - = = = + + - r r r r ˆ ˆ ˆ ˆ ˆ ˆ y x z inst x y z dv dv dv dv a i j k a i a j a k dt dt dt dt = = + + = + + r r Equations of Linear Motion in 1-D for Constant Linear Acceleration: Equ. (2-11): v x = v 0x + a x t Equ. (2-15): x - x 0 = v 0x t + ½a x t 2 Equ. (2-16): v x 2 = v 0x 2 + 2a x (x – x 0 ) Equ. (2-17): x - x 0 = ½(v 0x + v x )t Equ. (2-18) x - x 0 = v x t - ½a x t 2 x 0 , v 0x , and a x are constants. Note that Equ. (1) is the time derivative of Equ. (2). For linear motion in 2-D (e.g., projectile motion), you can treat the motions in the
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: x-direction and in the y-direction separately except that the time variable is the same for both motions. In problems involving projectile motion near the surface of the earth the acceleration in the y-direction is usually g=9.80 m/s 2 in the downward direction and the acceleration in the x-direction is usually zero because effects of air drag are ignored. At the highest point of its trajectory the vertical velocity component of a projectile is zero at the instant the projectile reverses its direction from going up to coming down. Phys 0174 – Fall 2008 – P. Koehler...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern