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# Physics 2110 Week 3 Notes PHYS 2110

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This 5 page Class Notes was uploaded by Mary Catherine Rush on Thursday September 8, 2016. The Class Notes belongs to PHYS 2110 at East Tennessee State University taught by Dr. Mark Giroux in Fall 2016. Since its upload, it has received 38 views. For similar materials see Technical Physics I in PHYSICS (PHY) at East Tennessee State University.

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Date Created: 09/08/16

PHYSICS2110NOTES Week three 4-1PositionandDisplacement When considering motion in 2-D and 3-D, you must find the particle’s position and displacement. Position of a particle can be found using a position vector ???? ⃗, which extends from a reference point (like the origin of a coordinate plane) to the particle In unit vector notation ???? ⃗ = x????̂ + y????̂ + z???? o Keep in mind that x????̂, y????̂, and z???? are vector components because they indicate magnitude and direction and x, y, and z are scalar components because they only indicate magnitude As a particle moves, the position vector changes as to always extend to the particle from the reference point. As this position changes, the particle’s displacement is described as Δ???? ⃗ = ????⃗2−???? 1 o Where the vectors ???? 2 and???? 1 can also be described in unit vector notation so that Δ⃗ = Δx????̂ + Δy????̂ + Δz???? or Δ????⃗ = (x 2x )????1 + (y −2 )????̂1+ (z − 2 )???? 1 ̂ 4-2AverageVelocityandInstantaneousVelocity We know that in 1-D, an objects average velocity is ???????????????????? ????????????????????????????????????????????????or ???????????????????? ???????????????? ????avg= ???????? and the instantaneous velocity is ???? = ???????? (the first derivative of ???????? ???????? position, x, with respect to time, t) However, as a particle moves in 2-D and 3-D through a displacement vector Δ???? ⃗ during a time interval ???????? then its average velocity as a vector quantity is ⃗ = Δ???? = Δx???? + Δy???? + Δz???? avg ???????? ???????? The instantaneous velocity ???? as a vector quantity is the limit of ???? ⃗avgas ???????? d????⃗ approaches 0 written as ???? =⃗⃗ (the first derivative of the position of ⃗ with ???????? respect to time, t) Graphically, the direction of the instantaneous velocity vector ???? of the particle is always tangent to the particle’s path at the particle’s position The magnitude and angle of the instantaneous velocity can be found using the same formulas as 1-D motion: ???????? ???? = √ ????????+ ???? an???? the angle ???? = ???????????? −1 ???????? 4-3AverageAccelerationandInstantaneousAcceleration We know that acceleration is the derivative of velocity, and average velocity is the particles change in distance over the total time traveled. So when a particle’s velocity changes, from1???? to2???? , the average acceleration is the ????2 − ????1 ????????⃗⃗ change in velocity over a time interval ????????, which ⃗avg= = ???????? ???????? And the limit of ????⃗avgas the time interval ???????? approaches 0 is the ????????⃗⃗ ̂ instantaneous acceleration, written a????⃗ = ???????? = ???? x̂ + ????y????̂ + ????z???? ???????????? ???????????? o The scalar components of ????⃗ can be rewritten as ???? =x ???????? ???? y ???????? ???????? ???? ????z= ???????? To find the magnitude and angle of????⃗ the same formulas from 1-D motion 2 2 −1 ???????? can be used. ???? = √ ????????+ ???? a????d the angle ???? = ???????????? ???? ???? 4-4ProjectileMotion When an object is launched at an angle, the object is moving in 2-D. For now, we are only worried about the movement of the object right after it was launched and right before it lands. This would make the initial velocity0 and its acceleration at a constant free- 2 fall,⃗, where ???? = 9.80 ????/???? o Air resistance is neglected as well as the Coriolis Effect (an effect the Earth’s rotation can have on a moving object) This type of particle, one that moves in a vertical plane with free-fall acceleration, is called a projectile and the motion of the particle is called projectile motion. The initial velocity can be written ⃗0= ???? ????0????+ ???? ????0???? Similar to the formulas used in 1-D motion, the x and y components of the initial velocity can be found if the angle0???? , which is between ⃗0and the positive x direction, is known. So ????0???? = ???? ????0???????? a0d ???? 0???? = ???? 0???????????? 0 In projectile motion, the vertical and horizontal motion are independent of one another so that one does not influence the other. During its flight, the projectiles position vecto⃗ and velocity vector ⃗ constantly change, but ????⃗ remains constant and is directed downward along the y-axis. There is no horizontal acceleration and the magnitude of the velocity vector ???? is constant, so ???? = ???? 0 However, along the y-axis, ???? ????s to the free-fall acceleration ⃗ = −????????̂, and ????????= 0 at the top of the trajectory. This means that after the projectile is launched the downward-directed acceleration causes the magnitude of the velocity to become less and less positive until it reaches 0 and then becomes negative. o The motion is symmetrical if the object lands at the same point it was launched: ???? = 0 where ???? = the range along the x-axis Remember that in 1-D motion, the displacement of an object along the x-axis could be described as ???? − ???? = ???? ???? + 1???????? and along the y-axis it was 0 0 2 described as ???? − ???? = ???? ???? + 1???????? 2 0 0 2 Because the horizontal and vertical motion are independent of each other, we can break up the motion of the object into a vertical component and a horizontal component. In 2-D motion, we must consider direction so the horizontal displacement is ???? − ???? 0 ???? ???? 0???? = 0 so we don’t have to worry about it). Because ???? 0????= ???? ????0???????? we0can say ???? − ???? = (????0???????????????? )0 0 The horizontal displacement is a little trickier. The vertical motion has a constant downward acceleration. This can be expressed using −????. 1 2 1 2 1 2 So ???? − ???? 0 ???? ???? 0???? 2(−????)???? = ???? ???? 0???? 2???????? = ???? ????????0???? ???? −0) 2???????? Now we can find the velocity of the y component which is ???? = ???? ???????????????? − ???????? ???? 0 0 Like earlier, the equations from 1-D movement can be applied to find an unknown When time is eliminated, we have ???? = ???????? ????????????????0) − 20(???? − ???? ) 0 1 1 So if we know ???? − ???? =0???? ???? 0???? ???????? = ???? ????????0???? ???? −0) ???????? , we can do some 2 2 algebra to get two equations: ???? = ???? − ???????? and ???? = (???? 2 − 2????(???? − ???? ) ???? 0???? ???? 0???? 0 We can set ???? 0 ???? = 0 ???? If we rearrange the first equation, we get ???? = ????0????????????0 ???? ???? ???? 2 Now, if we plug that in to equation 1, we get ???? = ???? ????0???????? 0 ???? ???????????????? − (2 ???? ???????????????? ) 0 0 0 0 and trig identities lead us to the equation of the projectile’s path, or trajectory: ????????2 ???? = ???????????????? ???? 0) 2 2(????0????????????????0) 2 o Note that this equation is in the form ???? = ???????? + ???????? which is the equation for a parabola. The trajectory is parabolic! The Range of a particle’s trajectory is the horizontal distance it travels until it reaches the initial height, or the height from which it was launched. The Range 1 2 can be written as: ???? = ???? ????0???????? ???? s0 0 = ???? ???????????????? 0 − ????????0) 2 2????0 When t is eliminated ???? = ????????????????0???????????????? 0 ???? Remember that 2???? = 2???????????????????????????????? is a trig identity. So the range can be ????0 rewritten as ???? = ????????????2???? 0 ???? We also know that the range of a sine function is -1 to 1, so R would reach a maximum when 2???? = 1=00° or ???? = 45°0

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