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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