Lecture 5: Vectors &

Motion in 2 Dimensions

Questions of Yesterday

2) I drop ball A and it hits the ground at t1. I throw ball B horizontally (v0y = 0) and it hits the ground at t2. Which is correct?

a) t1 < t2

b) t1 > t2

c) t1 = t2

Questions of Yesterday1) Can a vector A have a component greater than its

magnitude A?a) YESb) NO

2) What are the signs of the x- and y-components of A + B in this figure?

a) (x,y) = (+,+)b) (+,-)c) (-,+)d) (-,-)

A

B

Displacement in 2 Dimensions

Position vectors no longer accounted for by + and -

Displacement = change in position vector of object = r

1 Dimension

x = xf - xi r = rf - ri

2 Dimensions

x (m)

y (m)

ri

rf

r

object path

Velocity in 2 Dimensions

Average Velocity

rf - ri

tf - ti

rt

vav = =

rt

v = limt -> 0

Instantaneous Velocity

x (m)

y (m)

ri

rf

r

object path

ti

tf

Acceleration in 2 Dimensions

vf - vi

tf - ti

vt

aav = =

Instantaneous Acceleration

vt

a = limt -> 0

Average Acceleration

vx (m/s)

vy (m/s)

vi

vf

v

object’s instant. velocity

ti

tf

Acceleration in 2 Dimensions

vf - vi

tf - ti

vt

aav = =

If a car is going North at a constant speed and makes a left turn while maintaining its constant speed and

then continues West at the same speed…does the car accelerate during this trip?

Average Acceleration

A runner is running on a circular track at constant speed?

Is she accelerating?

Acceleration in 2 Dimensions

vf - vi

tf - ti

vt

aav = =

Velocity is a vector with both magnitude & direction, so…

An object can accelerate by eitherchanging its SPEED or

changing its DIRECTION

Average Acceleration

Projectile MotionMotion in 2 Dimensions under constant gravitational acceleration

Horizontal component of velocity is constant over entire path!vx = v0x

No acceleration in horizontal direction

Projectile MotionMotion in 2 Dimensions under constant gravitational acceleration

Vertical component of velocity constantly changingdue to gravitational acceleration in -y direction

v0y --> 0 -> -v0y

t= 1 s t= 2 s

t= 3 s

t= 4 s

t= 5 s

Projectile Motion

Horizontal and Vertical motions are completely independent of each other!!

Motion in one direction has NO EFFECT on motion in the other direction

Initial velocity in horizontal directionno gravity

Initial velocity in horizontal directionwith gravity

Initial velocity in vertical direction with gravity

Important Features of Projectile Motion

Acceleration is ALWAYS -9.80 m/s2 in the vertical direction

Parabolic motion is symmetricvf = -v0

At the top of the trajectory: t = 1/2 of total time

x = 1/2 of total horizontal range

t= 1 s t= 2 s

t= 3 s

t= 4 s

t= 5 s

Total time of trajectory is independent of horizontal motion

At what point in the object’s trajectory is the speed a minimum?

What about velocity?

t= 1 s t= 2 s

t= 3 s

t= 4 s

t= 5 s

Important Features of Projectile Motion

2D Motion under Constant Acceleration

Because x and y motions are independent…we can apply 1D equations for constant acceleration motion

separately to each x- and y- direction

But….v0 has both

x- and y-components

Need to separate v0 into x- and y- components

v0x = v0cosv0y = v0sin

2D Motion under Constant Acceleration

Recall equations for 1D motion under constant acceleration

v = v0 + atx = v0t + 1/2at2

v2 = v02 + 2ax

2D motion equivalent to superposition of two independent motions in the x- and y-directions

vx = v0x + axtx = v0xt + 1/2axt2

vx2 = v0x

2 + 2axx

Horizontal Motion

vy = v0y + ayty = v0yt + 1/2ayt2

vy2 = v0y

2 + 2ayy

Vertical Motion

Horizontal Motion of Projectile

vx = v0x + axtx = v0xt + 1/2axt2

vx2 = v0x

2 + 2axx

v0x = v0cos

ax = 0

Time is still determined by y-direction motion!

t= 1 s t= 2 s

t= 3 s

t= 4 s

t= 5 s

vx = v0cos = constantx = v0xt = (v0cost

Vertical Motion of Projectile

vy = v0y + ayty = v0yt + 1/2ayt2

vy2 = v0y

2 + 2ayy

v0y = v0sin

ay = g = -9.80 m/s2

vy = v0sin + gty = (v0sint + 1/2gt2

vy2 = (v0sin2 + 2gy

Pay attention to sign convention!

Equations for Motion of Projectile

vy = v0sin + gty = (v0sint + 1/2gt2

vy2 = (v0sin2 + 2gy

vx = v0cos = constantx = v0xt = (v0cost

v = (vx2 + vy

2)1/2

tan = vy/vx

= tan-1(vy/vx)-90 < < 90

Vertical Component

Motion

HorizontalComponent

Motion

Combined2D Motion

Problem #1A projectile falls beneath the straight-line path it would follow if there were no gravity. How many meters does it fall below this line if it has been

traveling for 1 s? For 2 s?

Does your answer depend on the angle at which the projectile is launched? What about the speed?

t= 1 s t= 2 s

t= 3 s

t= 4 s

t= 5 s

Questions of the Day2) Two projectiles are thrown with the same initial

speed, one at an angle with respect to the ground and the other at an angle 90o - . Both projectiles strike the ground at the same distance from the projection point. Are both projectiles in the air for the same length of time?

a) YESb) NO

1) A heavy crate is dropped from a high-flying airplane as it flies directly over your shiny new car? Will your car get totaled?

a) YESb) NO