Kinematics in two dimensions Position Displacement Average Velocity (Instantaneous) Velocity Average Acceleration (Instantaneous) Acceleration.

Kinematics in two dimensions

• Position

• Displacement

• Average Velocity

• (Instantaneous) Velocity

• Average Acceleration

• (Instantaneous) Acceleration

ˆ ˆr xi yj

2 1ˆ ˆr r r xi yj

rv

t

drv

dt

va

t

dva

dt

Instantaneous velocity and acceleration

x y z

dr dx dy dzˆ ˆ ˆ ˆ ˆ ˆv i j k v i v j v kdt dt dt dt

yx zx y z

dvdv dvdv ˆ ˆ ˆ ˆ ˆ ˆa i j k a i a j a kdt dt dt dt

Constant acceleration

0v v at 2

0 0

1x x v t at

2

1 - D

0v v at

20 0

1r r v t at

2

3 - D

x x0 xv v a t

20 x0 x

1x x v t a t

2

2 2x x0 x 0v v 2a x x

y y0 yv v a t

20 y0 y

1y y v t a t

2

2 2y y0 y 0v v 2a y y

0

0

0

ya g

Projectile Motionx x0v v

0 x0x x v t

y y0v v gt

20 y0

1y y v t gt

2

2 2y y0 0v v 2g y y

Mortars, artillery, and guns

x0 0v v cos

y0 0v v sin

2 20 0

m

v siny

2g

2

0 0v sin2R

g

y0 22

x0 x0

v gy x x

v 2v

Motion is parabolic!

2004 Report on the status of the Army Artillery School

• ...And student officers with reading and math skills below seventh-grade level have been allowed to recycle multiple times before being force-fed to their units - even though an artillery officer who can't do math is like a fireman who's allergic to smoke.

• Columnist and former soldier

David H. Hackworth is the author of The Price of Honor, and contributes weekly commentary to DefenseWatch

Example Problem

• A batter hits a pitched ball whose center is 4.0 feet above the ground so that is angle of projection is 45 degrees and its range is 350 feet. The ball will be a home run if it clears a 24 ft high fence that is 320 feet from home plate. Will the ball clear the fence? If so, by how much?

Uniform Circular Motion

• Uniform = magnitude of velocity (speed) is not changing

2

R

va

r

Uniform Circular Motion Terms

• Frequency (f) – The number of revolutions (cycles) completed in a second.

• Period (T) – The time to complete one revolution (cycle).

1T

f

distance in one revolution 2 rv

time to complete one revolution T

Uniform Circular Motion Example

• A particle moves in a circular path 0.40 m in radius with constant speed. The particle makes 5.0 revolutions in each second of its motion.

• Find the particle’s speed

• Find the particle’s acceleration

13 m/s

390 m/s2

Relative Motion

BS BW WSv v v

The velocity of an object relative to stationary reference frame is equal to the velocity of an object relative to a moving reference frame plus the velocity of the moving reference frame.

Relative Motion

There is a 3 knot westerly current. Your ship is doing 10 knots by log. What course must you steer to make course north good?What will be your speed over ground?

017

9.5 kts

Mo Boards

0

3 min

1000 yds

090/15kts

135/10kts

BS B you you Sv v v