**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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