Displacement, velocity and acceleration vectors Projectile ...
[Pages:16]Ch04: Motion in two and three dimensions (2D and 3D)
Displacement, velocity and acceleration vectors
Projectile motion Circular motion Relative motion
4.2: Position and displacement
Position of an object in 2D or 3D is described by its position vector
is drawn from origin to the position of the object at a given time. It can be written as
Position vector in 3D
As particle position changed from
during a certain
time interval ? displacement vector that can be written as
In components form ?
4.2: Position and displacement
Displacement vector in 2D is shown in the figure below:
rr = rr2 - rr1
rr = xi^ + y^j x = x2 - x1 y = y2 - y1
4.2: Position and displacement: Example1
Rabbit runs across a parking lot. Its position changing according to x and y in meters, t in seconds. Find its position vector at t=15s in unit vector notation and in magnitude angle notation
? Position vector in unit vectors can be written as
The magnitude from the origin is
at angle
4.3: Average velocity and Instantaneous Velocity
Since occurs in time
interval t = t2 - t1
? average velocity
4.3: Average velocity and Instantaneous Velocity
If t decrease until t?0 ? We will have Instantaneous velocity (or velocity)
The Speed v = vr =
vx2
+
v
2 y
4.3: Average velocity and Instantaneous
Velocity: Example 2
From Ex:1
For the rabbit in the preceding example1,
find the velocity at time t =15 s.
Velocity vector is
4.4: Average acceleration and Instantaneous acceleration
When a particle's velocity changes from
in a time
interval , its average acceleration during is
Note that, for an object, acceleration can cause a change in the velocity magnitude or in the velocity direction, or both.
4.4: Average acceleration and Instantaneous acceleration
For ?0 we will have Instantaneous acceleration (or acceleration)
4.4: Average acceleration and Instantaneous acceleration: Example 3
For the
the rabbit in acceleration
tarheapt rtiemceedtin=g15twso.
examples
1
and
2,
find
From Ex: 2 we have
and Hence the acceleration vector is
4.4: Average acceleration and Instantaneous acceleration: Example 4
Trrh(et
position of a
) = 2t 2i^ + 3t^j
particle where
moving in the xy-
r is in meter and
plane is given t is in seconds.
by Find
a) the average velocity in the time interval t = 2s and t = 4s
vravg
=
rr t
=
rr2 t1
- rr1 - t1
=
rr(4) - rr(2) 4-2
=
2(42 )i^ + 3(4) ^j - 2(22 )i^ - 3(2) ^j 4-2
= (32 - 8)i^ + (12 - 6) ^j = (12i^ + 3 ^j) m / s 2
b)
Vvr(etl)o=cidtyrra=n4dti^sp+e3e^jdofvrt(h3e)
particle at t = 3s = 4(3)i^ + 3 ^j = 12i^ +
3
^j
m
/
s
dt
c)
speed = vr = 122 The acceleration
+ 32 = at t=3s
153 ?
ar==12d.v3r 7=m4i/^
s m
/
s2
at any time (ar is constant)
dt
4.4: Average acceleration and Instantaneous acceleration: Example 5
A particle starts from the origin at t = 0 with an initial velocity having an x component of 20 m/s and a y component of -15 m/s. The particle moves in the xy plane with an acceleration in the x direction only (ax= 4.0 m/s2). Find
a) The components of the velocity vector at any time and the total velocity vector at any time.
vx = v0x + axt = ( 20 + 4t)m/s vy = v0 y = -15m/s (ay = 0 ) velocity vector vrf = vxi^ + vy ^j = ((20 + 4t)i^ -15 ^j)m/s
b) Calculate the velocity and speed of the particle at t = 5 s.
vr(t=5) = (20 + 4(5))i^ -15 ^j = (40i^ -15 ^j)m / s speed = vr(t=5) = 402 + (-15)2 = 1825 = 42.7 m / s
4.5: Projectile Motion
Projectile motion is an example of motion in 2D Consider the Two assumptions:
1) vTehleocoitbyjevrc0t,haansdains initial
launched at an angle 0 2) Free-fall acceleration (g)
is constant (neglect air resistance)
Then the trajectory (path) of the projectile is a parabola. The velocity changes magnitude and direction The acceleration in the y-direction is constant (-g). The acceleration in the x-direction is zero.
4.5: Projectile Motion
The initial velocity vector of projectile motion is where
4.6: Projectile Motion analysis
We'll analyze projectile motion as a superposition of two independent motions (x and y directions):
Constant velocity motion in the horizontal direction (x)
Free-fall motion in the vertical direction (y)
vx = v0x = v0cos0 = constant
vy = v0 y - gt
x - x0 = x = vxt = v0xt
y
-
y0
=
y
=
v0 yt
-
1 2
gt 2
From x and y (where x0 and y0 = 0 at t = 0) ? we eliminate t ? equation of the projectile
path is
4.6: Projectile Motion analysis: the maximum height and horizontal range
Ex: Assume the projectile is launched with initial speed v0 at angle 0, if the maximum hight is h and the horizontal range is R. Determine
a) h and b) R
a) h is in y-axis ? ay = ?g ;to find h we use
h
=
y
=
v0
yt
A
-
1 2
gt
2 A
( t
A
;
time
to
reach
max.
hight
h)
v0
tA can be found using
0
vy = vyA = v0 y - gtA (at max. hight, vyA = 0)
0
=
v yi
-
gt A
tA
=
v0 y g
But
v0 y
=
v0
sin 0
tA
=
v0
sin 0 g
hence
h
=
v0
sin 0
v0
sin 0
g
-
1 2
g
v0
sin g
0
2
h = v02 sin 2 0
2g
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