Chapter 4, Motion in 2 Dimensions

Position, Velocity, Acceleration• Just as in 1d, in 2d, an object’s motion is

completely known if it’s position, velocity, & acceleration are known.

• Position Vector r– In terms of unit vectors discussed last time,

for an object at position (x,y) in x-y plane:

r x i + y j For an object moving: r depends on time t:

r = r(t) = x(t) i + y(t) j

• Suppose that an object moves from point A(ri) to point B(rf) in the x-y plane:

• The DisplacementVector is: Δr = rf - ri

If this happens in in a time Δt = tf - ti

• The Average Velocity is: vavg (Δr/Δt)

Obviously, this in the same direction as the displacement.

It is independent of the path between A & B

• As Δt gets smaller & smaller, clearly, A & B get closer & closer together. Just as in 1d, we define

The Instantaneous Velocity Velocity at Any Instant of Time.

average velocity over an infinitesimally short time• Mathematically, the instantaneous velocity is:

v lim∆t 0 [(∆r)/(∆t)] ≡ (dr/dt)lim ∆t 0 ratio (∆r)/(∆t) for smaller & smaller

∆t. Mathematicians call this a derivative. The instantaneous velocity

v ≡ time derivative of displacement r

• Instantaneous Velocity v ≡ (dr/dt). • The magnitude |v| of vector v ≡ speed. As motion

progresses, the speed & direction of v can both change. For an object moving from A (vi) to B (vf) in the x-y plane: Velocity Change Δv = vf - vi

This happens in time Δt = tf - ti

• Average Acceleration aavg (Δv/Δt) As both the speed & direction of v change, over an arbitrary path

• As Δt gets smaller & smaller, clearly, A & B get closer & closer together. Just as in 1d, we define

The Instantaneous Acceleration Acceleration at Any Instant of Time. average acceleration over infinitesimally short time

• Mathematically, instantaneous acceleration is:

a ≡ lim∆t 0 [(∆v)/(∆t)] ≡ (dv/dt)lim ∆t 0 ratio (∆v)/(∆t) for smaller & smaller ∆t. Mathematicians call this a derivative.

The instantaneous acceleration a ≡ time derivative of velocity v

2d Motion, Constant Acceleration• It can be shown that:

Motion in the x-y plane canbe treated as 2 independent

motions in the x & y directions. So, motion in the x direction

doesn’t affect the y motion & motion in the y direction doesn’t

affect the x motion.

• Suppose an object moves from A (ri,vi), to B (rf,vf), in the

x-y plane. So, it’s position clearly changes with time: The acceleration a is constant, so, as in 1d, we can

write (vectors!): rf = ri + vit + (½)at2

• If its velocity also changes with time, it is accelerating. If it’s acceleration a is constant, so, as in 1d, we

can write (vectors!): vf = vi + at

Summary• For 2d motion when the acceleration a is constant, we have:

rf = ri + vit + (½)at2, vf = vi + at• For the Horizontal Part of the Motion we have:

xf = xi + vxit + (½)axt2, vxf = vxi + axt• For the Vertical Part of the Motion we have:

yf = yyi + vyit + (½)ayt2, vyf = vyi + ayt

Projectile Motion

A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola.

Projectile Motion• Projectile Motion Motion of an object

that is projected into the air at an angle.• Near the Earth’s surface, the acceleration a

on the projectile is downward and equal toa = g = 9.8 m/s2

Goal: Describe the motion after it starts.• Galileo: Analyzed horizontal & vertical

components of motion separately.• Today: Displacement D & velocity v are vectors

Components of the motion canbe treated separately

Equations Needed• One dimensional, constant acceleration equations

for x & y separately!

x part: Acceleration ax = 0! y part: Acceleration ay = g!

(+ if down is positive - if down is negative).Take down positive for a while!

• The initial x & y components of the velocity: vxi & vyi.

x motion: vxf = vxi = constant. xf = xi + vxi t

y motion: vyf = vyf + gt, yf = yi + vyi t + (½)g t2

(vyf) 2 = (vyi)2 + 2g (yf - yi)

Projectile Motion• Simplest Example: A ball rolls across the table, to the

edge & falls off edge to floor. It leaves the table at time t = 0. Analyze the x & y parts of the motion

separately.• y Part of the Motion: Down is positive. The origin is at

the table top: yi = 0. Initially, there is no y component of velocity: vyi = 0

vyf = gt, yf = (½)g t2

• x Part of the Motion: The origin is at the table top: xi = 0.

There is NO x component of acceleration(!) ax = 0. Initially, the x component of velocity is: vxi

vxf = vxi , xf = vxit

Can be understood by analyzing the horizontal & vertical motions separately. Take down as positive. The initial velocity has an x component ONLY!

That is vyi = 0.

The Ball Rolls Across the Table & Falls Off

At any point, v has both x & y components. The kinematic equations tell us that, at time t,

vx = vxi, vy = gtxf = vxit

yf = (½)gt2

vxi

t = 0 here

Summary• A ball rolling across the table & falling.

• The vector velocity v has2 Components:

vxf = vxi , vyf = gt

• The vector displacement D also has2 components:

xf = vxit , yf = (½)gt2

The speed in the x-direction is constant; in the y-direction the object moves with constant acceleration g.

The photo shows two balls that start to fall at the same time. The one on the right has an initial speed in the x-direction. It can be seen that vertical positions of the two balls are identical at identical times, while the horizontal position of the yellow ball increases linearly.

The PHYSICS of Projectile Motion

The vertical (y) part of the motion:vyf = gt , yf = (½)g t2

Is the SAME as free fall!!

The horizontal (x) part of the motion:vxf = vxi, xf = vxit

Is motion at constant velocity!! An object projected horizontally will reach the

ground at the same time as an object dropped vertically from the same point!

(the x & y motions are independent)

General Case: An object is launched at initial angle θ0 with the horizontal. The analysis is similar to before, except the initial velocity has a vertical component vy0 0. Let up be positive now!

The acceleration = g downward for the

entire motion!

The Parabolic shape of the path is real (neglecting air resistance!)

vx0 = v0cosθ0

vy0 = v0sinθ0

Projectile Motion Diagram

• General Case: Take y positive upward & origin at the point where it is shot: xi = yi = 0

vxi = vicosθi, vyi = visinθi

• Horizontal motion: NO ACCELERATION IN THE x DIRECTION!

vxf = vxi , xf = vxi t

• Vertical motion:

vyf = vyi - gt , yf = vyi t - (½)gt2

(vyf) 2 = (vyi)2 - 2gyf

– If y is positive downward, the - signs become + signs.

ax = 0, ay = -g = -9.8 m/s2

Summary: Projectile MotionProjectile motion is motion with constant acceleration in two dimensions, where

The acceleration is g and is down.

Projectile Motion: More Discussion• As we already said, the y-component of the velocity

is zero at the maximum height of the trajectory• The acceleration stays the same throughout the trajectory

Range & Maximum Height• When analyzing

projectile motion, two characteristics are of special interest:

• The Range, R = the horizontal distance of the projectile

• The Maximum Height h = the highest point.

Maximum Height• The Maximum Height h of a projectile is the height

at which the y component of velocity vy = 0. • It can be obtained from the kinematic equation:

(vyf) 2 = (vyi)2 + 2g (yf - yi)

• Assuming yi = 0, setting vyf = 0 & putting yf = h: 0 = (vyi)2 + 2gh

• So: h = [(vyi)2/(2g)]

• Using vyi = vi sinθi gives:

• Note that this is valid only for symmetric motion

2 2sin

2i iv

hg

Range• The Range of a Projectile, R = xmax, is the maximum

horizontal distance from the starting point. • It can be obtained from the kinematic equation

R = xmax = vxitmax (1)with tmax = the time for it to hit the ground.

• The time to hit tmax can be obtained by setting yf = 0 in the kinematic equation yf = vyi t - (½)gt2. This gives

tmax = (2vyi/g) (2)• Putting (2) into (1) gives

R = (2vxivyi/g) (3)• In (3), using vxi = vicosθi & vyi = visinθi along with the trig identity 2

sinθicosθi = sin(2θi) gives the final result:

• Note that this is valid only for a symmetric trajectory

2 sin2i ivR

g

Ranges for Fixed vi & Changing θi

Range

The Range is:2 sin2i iv

Rg

• Using differential calculus, setting (dR/di) = 0 & solving for i gives the result that

The maximum range R occurs at i = 45o,

independent of the initial velocity vi.

• Complementary angles will produce the same range

– The maximum height will be different for the two angles

– The times of the flight will be different for the two angles

Non-Symmetric Projectile Motion• Follow the general rules

for projectile motion

• One way is to break the y-direction into parts– up and down or – symmetrical back to

initial height and then the rest of the height

• Apply the problem solving process to determine & solve the necessary equations