3.Motion_2or3D

# In velocit y time interval y 0 22 t 2 9 1t 30 rrr

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Unformatted text preview: v1 r a avg = = ∆t t 2 − t1 with t in second and x and y in meters. 1) At t = 15s, what is the rabbit’s position vector in unit vector notation and magnitude-angle notion? 2) What is the velocity vector in unit vector notation and magnitude-angle notation? The Department of Physics, CUHK The Department of Physics, CUHK 9 Instantaneous acceleration 10 Instantaneous acceleration If we shrink the time interval to zero about some instant, the average acceleration approaches the instantaneous acceleration at that instant, i.e. r ˆ a = axˆ + a yˆ + az k i j r r dv a= dt ax = In unit-vector notation, we can rewrite it as dv dv dv ˆ rd ˆ a = (v x ˆ + v y ˆ + v z k ) = x ˆ + y ˆ + z k i j i j dt dt dt dt dv y dv x dv , ay = , az = z dt dt dt Attention: if the velocity changes in either magnitude or direction (or both), the particle must have an acceleration. The Department of Physics, CUHK The Department of Physics, CUHK 11 12 2 §5. Projectile motion Sample problem 4 When a particle moves in a vertical plane with some initial velocity, but its acceleration is always the free-fall acceleration, which is downward, the particle’s motion is call projectile projectile motion. The rabbit runs across a parking on which a set of coordinate axes has, strangely enough, been drawn. The coordinate of the rabbit’s position as functions of time t are given by x = − 0 .31t 2 + 7 .2t + 28 , 柯受良飞越黄河 y = 0 .22 t 2 − 9 .1t + 30 Aristotel’s cannon with t in second and x and y in meters. 3) What is the acceleration vector in unit vector notation and magnitude-angle notation? The Department of Physics, CUHK The Department of Physics, CUHK 13 14 §6. Projectile motion analyzed Projectile motion projectile In projectile motion, the horizontal motion and the vertical motion are independent of each other, neither motion effects the other. r v = v 0 x ˆ + v 0 y ˆj i v 0 x = v cos θ 0 The horizontal motion: x − x0 = v0 x t x − x 0 = (v 0 cos θ 0 )t v 0 y = v sin θ 0 The vertical motion: y − y 0 = v 0 y t − 1 gt 2 2 = ( v 0 sin θ 0 )t − 1 gt 2 2 v y = v0 sin θ 0 − gt The Department of Physics, CUHK The Department of Physics, CUHK 15 Projectile motion analyzed Sample problem 7 The following figure shows a pirate ship 560 m from a fort defending the harbor entrance of an island. A defense cannon, located at sea level, fires balls at initial speed v0=82m/s. The equation of the path gx 2 y = (tan θ 0 ) x − 2 ( v0 cos θ 0 ) 2 The horizontal range a) At what angle from the horizontal must a ball be fired to hit the ship. y − y0 = 0 ⎧ R = ( v0 cos θ 0 )t ⎨ 2 1 ⎩0 = ( v 0 sin θ 0 )t − 2 gt 16 R= 2 v0 sin 2θ 0 g b) How far should the pirate ship be from the cannon if it is to be beyond the maximum range of the cannonballs. This equation only gives the horizontal distance when the final height is the same as the launch height. The Department of Physics, CUHK The Department of Physics, CUHK 17 18 3 §7. Uniform circular...
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## This document was uploaded on 09/16/2013.

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