Motion in 2 and 3 DimensionsThe position of a particle in space, relative to a given origin, can be represented at any given time by the vector r, where r = rxi + ryj + rzk.

Since the object can have component displacements in 2 or 3 dimensions, it can have component velocities and accelerations:vx = dx vy = dy vz = dz

dt dt dt

and

ax = dvx ay = dvy az = dvz

dt dt dt

A ball is thrown upward with a velocity of 35.0 m/s. A wind from the left imparts a constant horizontal acceleration of 1.10 m/s2 on the ball. When the ball has reached its highest point, find A) how far right it has drifted from the point at which it was thrown, and B) what is the speed of the ball?

vyo = 35.0 m/svy = 0ay = -g

vxo = 0

ax = 1.10 m/s2

v = ?

x = ?

x = vxot + .5axt2 vy = vyo + ayt

t = 3.57 sx = 7.01 m

v = √ vx2 + vy

2 vy = 0

vx = vxo + axt

v = 3.93 m/s

v = 3.93i + 0j m/s

A skier moves down a flat 10° incline. A lateral wind gives him a lateral acceleration of .54 m/s2. If he pushed off with a velocity of 9.0 m/s down the slope and no lateral

component to his velocity. If the frictionless slope is 125 m long and 25 m wide, find A) where he leaves the slope and B) what is his velocity at that point.

Projectile Motion

A specific type of two dimensional motion where the constant acceleration of the object is - g!

• if the object is a projectile, then ax = 0

• therefore, the x motion of the object is constant!

• note that whether the object in freefall is rising, falling, or at the top of its trajectory, its acceleration is a constant - g!

A plane flying with a constant horizontal speed of 155 km/h and an elevation of 225 m wishes to drop a package dead center onto a target it is going to fly directly above. What sight angle should the package be released at?

ø

x

y

tan ø = x /y

With the origin at the release point:

vyo = 0

vxo = 155 km/h= 43.1 m/s

∆y = - 225 m

∆x = ?

∆x = vxt

t = √2∆y/g

ay = - g= 6.78 s

∆x = (43.1)(6.78)

= 292 m

ø = tan-1 (292 / 225)= 52.4˚

A soccer player kicks a ball with a velocity of 15.5 m/s directed 36˚ above the horizontal. Assuming the ball moves in a vertical plane, find A) how long it takes the ball to reach its maximum height B) its maximum height C) its range D) its impact velocity.

vyo = vosinø = 9.1 m/svo = 15.5 m/s, 36˚

vxo = vx = vocosø = 12.5 m/s

A) at max y: vy = 0

t = vy - vyo

a

= .93s

B) ymax = ? ∆y = vyot + .5at2

= (9.1)(.93) + .5(-9.8)(.93)2 = 4.2 m

C) t = .93(2) = 1.86s

∆x = vxt = (12.5)(1.86) = 23.3 m

D) v = √ vx2 + vy

2 vy = vyo + at= 9.1 + (-9.8)(1.86) = - 9.1 m/s

v = √ (12.5)2 + (-9.1)2 = 15.5 m/s

ø = tan-1 (vy / vx) = tan-1(- 9.1/ 12.5) = - 36.0˚

A rocket is launched from rest and moves in a straight line at 70.0˚ above the horizontal with an acceleration of 46.0 m/s2. After 30.0 s of powered flight, the engines shut off and the rocket follows a parabolic path back to the Earth. A)Find the time of flight from launch to impact. B)What is the maximum altitude reached? C)What is the horizontal distance from launch pad to the impact point?

A box slides down a steep roof from rest and accelerates at (3.85i − 4.60j) m/s2 for 5.00 s at which point it leaves the roof and becomes a projectile. It then lands 11.8 m horizontally away from the edge of the roof. How high is the edge of the roof above the ground?

Uniform Circular Motion

An object that travels in a circular path at constant speed:

• speed is constant, but not velocity because direction continuously changes

• acceleration is constant in magnitude and direction (radially towards the center).

• displacement is generally discussed in terms of radians (radius lengths) which is

r = d

v1r

v2

r d = v∆t

v1v2

∆v

v1 = v2 = v

∆v = v∆t

v r

∆v = v2

∆t r

a = v2 / r

Relative Motion

The observed motion of a single particle by two different observers that are in uniform motion relative to each other.

• to discuss relative motion we must realize perspectives other than our own:• If you are traveling in a car at 35 m/s, the speed of you relative to the ground is 35 m/s.• The speed of the ground relative to you could be viewed as - 35 m/s (considering yourself “stationary”).• reference points are arbitrary!

A plane flies with an airspeed of 215 km/h heading due east while a steady wind blows 65 km/h due north. What is the velocity of the plane with respect to the ground?

vPA

vAG

vPG

Ø

vPG = vPA + vAG

vPG = √ vPA2 + vAG

2

= √(215)2 + (65)2

= 225 km/h

ø = tan-1 (vAG)/(vPA) = tan-1(65)/(215) = 16.8˚

During a hailstorm, hail falls straight down to the earth at a speed of 8.5 m/s. What is the velocity of the hail relative to a driver traveling a straight road at 61 km/h?

Use driver’s perspective!vHD = vHG + vGD

vHG

vGD

vHD ø

vHD = √ (-16.9)2 + (8.5)2 = 18.9 m/s

vDG = 16.9 m/s

ø = tan-1 (vgd)/(vhg) = 63.3˚ off of the vertical

vGD = - 16.9 m/s

Object A moves with a velocity of 25.0 m/s at 35.0 ˚ while object B moves with a velocity of 45.0 m/s at an angle of 210˚. Find A) the velocity of A with respect to B and B) the velocity of B with respect to A.

vAG = 25.0 m/s @ 35.0˚

vBG = 45.0 m/s @ 210˚

vAB = vAG + vGB

vGB = 45.0 m/s @ 30˚