physics 203 college physics i fall 2012

Physics 203 – College Physics I Department of Physics – The Citadel

Physics 203College Physics I

Fall 2012S. A. Yost

Chapter 3Motion in 2 Dimensions – Part 2


Today’s Topics1. Motion Concepts in 2 Dimensions

We will use vectors to define displacement and velocity in 2 dimensions.

2. Constant Acceleration: Projectile Motion


Tuesday’s Assignment

Read Ch. 4, sec. 1 – 5 (Newton’s Laws) .

A problem set on HW3 on Ch. 3 is due Tuesday.

The first exam is scheduled for next Thursday.

You do not need to memorize equations: the essential ones will be provided for the exam.

We will finish Ch. 4 after the exam. The exam only covers chapters 1 – 3.


Question 4 from the Quiz…Vector A has a magnitude of 10 and a direction angle

θ = 60o measured counter-clockwise from the +x axis. What are the magnitude and direction angle of the vector – 2A?

A. – 20, 60o

B. 20, 240o

C. 20, – 30o

D. – 20, 240o

E. – 20, – 30o

→

→

x

y A→

θ = 60o


Question 4 from the Quiz…Vector A has a magnitude of 10 and a direction angle

θ = 60o measured counter-clockwise from the +x axis. What are the magnitude and direction angle of the vector – 2A?

A. – 20, 60o

B. 20, 240o

C. 20, – 30o

D. – 20, 240o

E. – 20, – 30o

→

→

x

yA→

θ = 60o

– 2A→

θ = 240o

2010


Two Ships

Suppose one ship (A) is 20o E of N from a port, at a distance of 2300 m. A second ship (B) is 50o W of N, 1100 m from the port.

Sketch the two position vectors of the ships and the displacement vector from ship A to ship B.

Calculate the magnitude and direction of the displacement vector from ship A to ship B.


Ships’ Position Vectors

x

yN

0 EW

S

Cartesian coordinate system

Ship 1 = AMagnitude A = 2300 m

Ship 2 = BMagnitude B = 1100 m

20o

2300 m

A→

→

→

50o

1100 m

B→


Motion in 2 DimensionsWe want to describe how

an object moves along a general path. . .

• where it is at any time

• how fast it moves and which direction.

DisplacementVelocityAcceleration

All vectors!


Motion and DisplacementsWhen an object

moves, the vector r which describes its position changes with time.

→

t = 0 st = 1 st = 2 st = 3 st = 4 st = 5 st = 6 st = 7 st = 8 s

O

r→


Motion and Displacements

The change in position from start to finish is the displacement vector

Δr = rf – ri.

The displacement only depends on the initial and final points, not the path.

Δ r

O

→ →→ →

→

→rf

ri


Average Velocity

The average velocity for the motion is the vector v defined to be the net displacement vector divided by the time:

v = Δ r / Δ t

Δ rv→

→

→→

→


Average Velocity

The average velocity always points in the same direction as the displacement. Δ r

v→→


SpeedThe rate at which the

particle moves along the path is its instantaneous speed.

Speed is a scalar, and is always positive.

t = 0 st = 1 st = 2 st = 3 st = 4 st = 5 st = 6 st = 7 st = 8 s

O

r→


Average Speed

The average speed means something different from average velocity:

The average speed is the total distance traveled along the path divided by the time:

v = l / t.

l

_


Instantaneous Velocity

The instantaneous velocity is a vector whose magnitude is the instantaneous speed, and whose direction is the direction of motion at that moment.

v

Instantaneous speed v = |v|→


Constant Velocity

The velocity is constant when the object moves in a straight line at a constant speed.

O

t = 0 st = 1 st = 2 st = 3 st = 4 st = 5 s r→


Constant Velocity vs Constant Speed

Both of our examples so far had constant speed, meaning that the particle follows the path at a uniform rate.

But only the motion in a straight line had constant velocity.

The velocity is constant only when the magnitude and direction are both fixed.

If the velocity vector changes, the motion has acceleration. Acceleration is also a vector, but we will consider only special cases.


Acceleration

The acceleration gives the rate of change of the velocity vector, and is also a vector.

The acceleration is zero only when an object is at rest or moves in a straight line at a constant speed.

If the acceleration is constant, that means the velocity only changes in the direction the acceleration vector points, and it changes at a constant rate.


Projectile MotionProjectile motion is

the motion of an object under the influence of gravity alone.

Constant velocity in the x direction,

Constant acceleration in the y direction (g = 9.80 m/s2)

Acceleration vector g points downward with magnitude g.

→


Path of a Projectile


Independence of Motions

The vertical and horizontal motion are independent.

The yellow ball is initially given a horizontal velocity, while the red one is just dropped.

They reach the same height at every time, because they have identical vertical motion.


Projectile Motion

A projectile is shot from the edge of a cliff as shown with an initial speed v0 = 65.0 m/s. The height of the cliff is h = 125 m, and the angle with horizontal is q = 37.0o.

a) What velocity does it have just before it hits the ground? (magnitude and direction)

y

x0

(b) How long does it take to hit the ground?


Football

A football is kicked from the 30 yard line with an initial angle of 35o. What is the minimum initial speed of a kick at this angle that can make it over the goal posts to score a field goal?

(The distance of the kick is 40 yards = 120 ft, and the goal post bar is 10 ft high.)

x

yv0

d

h

→

θ


Football

Given: d = 120 ft, h = 10 ft, θ = 35o.

Horizontal: d = v0x t with v0x = v0 cos θ

Vertical: h = v0y t – ½ g t2 with v0y = v0 sin θ

x

yv0

d

h

→

θ


FootballSubstitute t = d/v0x into h = v0y t – ½ g t2 :

h = d (v0y/v0x) – ½ g d2/v0x2

= d tan θ – ½ g (d2/v02) sec2 θ

Simplify:

2 cos2 θ (d tan θ – h) = ½g d2/v02

d/v0 = 2cos θ √(d tan θ – h)/g = 1.638 √ 7.55 s2

120 ft/v0 = 4.50 s

v0 = 120 ft / 4.50 s = 26.7 ft/s

physics 203 college physics i fall 2012

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