Chapter 3 Motion in 2 dimensions. 1) Displacement, velocity and acceleration displacement is the vector from initial to final position.

Chapter 3Motion in 2 dimensions

1) Displacement, velocity and acceleration

• displacement is the vector from initial to final position

Δrr =

rr −

rr0

Δrx = x - component of Δrr = Δx

Δry = y - component of Δrr = Δy

Δy

Δx

Δx = rx − r0 x = x − x0

Δy = ry − r0 y = y − y0

1) Displacement, velocity and acceleration

• average velocity

rv =

Δrr

Δt

rv

vx =ΔxΔt

=x−x0

Δt

vy =ΔyΔt

=y−y0Δt

1) Displacement, velocity and acceleration

• instantaneous velocity

rv = lim

Δt→ 0

Δrr

Δt

vx = limΔt→ 0

ΔxΔt

vy = limΔt→ 0

ΔyΔt

rv0

rv

v can change even if v is constant

v is tangent to the path

1) Displacement, velocity and acceleration

• average acceleration

ra =

Δrv

Δt=

rv−

rv0

Δt

rv0

rv

rv

ra

ax =Δvx

Δt

ay =Δvy

Δt

not in general parallelto velocity

1) Displacement, velocity and acceleration

• instantaneous acceleration

rv0

rv

ax = limΔt→ 0

Δvx

Δt

ay = limΔt→ 0

Δvy

Δt

ra = lim

Δt→ 0

Δrv

Δt

1) Displacement, velocity and acceleration

• instantaneous acceleration

rv0

rv

ra = lim

Δt→ 0

Δrv

Δt

- object speeding up in a straight line

rv0

rv

Δrv

acceleration parallel to velocity

- object at constant speed but changing direction

rv Δ

rv

acceleration perp. to velocity

ra

2) Equations of kinematics in 2d

• Superposition (Galileo): If an object is subjected to two separate influences, each producing a characteristic type of motion, it responds to each without modifying its response to the other.

• That is, we consider x and y motion separately

2) Equations of kinematics in 2d

A bullet fired vertically in a car moving with constant velocity, in the absence of air resistance (and ignoringCoriolis forces and the curvature of the earth), will fall back into the barrel of the gun. That is, the bullet’s x-velocity is not affected by the acceleration in the y-direction.

vx

vy

vy

vx

2) Equations of kinematics in 2d

• That is, we can consider x and y motion separately

2) Equations of kinematics in 2d

• That is, we can consider x and y motion separatelyDisplacement : x, y (x0 =0,y0 =0)Velocity: vx,vy v0x,v0y

Acceleration: ax,ay (a0x =ax,a0y =ay)

Time: t (t0 =0)

x =v0xt+12 axt

2

vx =v0x + axt

vx2 =v0x

2 + 2axx

x= 12 (v0x + vx)t

x=vxt−12 axt

2

y =v0yt+12 ayt

2

vy =v0y + ayt

vy2 =v0y

2 + 2ayy

y= 12 (v0y + vy)t

y=vyt−12 ayt

2

Example

rv0

ra

x

y v0 x =22 m/s; v0y =14 m/s

ax = 24 m/s2; ay =12 m/s2

t=7.0 s

Find x, y,rv at t =7.0 s

x =v0xt+12 axt

2 =740 m

y=v0yt+12 ayt

2 =390 m

vx =v0x + axt=190 m/svy =v0y + ayt= 98 m/s

v = vx2 + vy

2 = 210 m/s

tanθ =vy

vx

=0.516 → θ =27º

rv

v0 = v0x2 + v0y

2 = 26 m/s

tanθ =vy

vx

=0.516 → θ =32.5º

rv

3) Projectile Motion (no friction)

a) EquationsConsider horizontal (x) and vertical (y) motion separately (but with the same time)

Horizontal motion: No acceleration ==> ax=0

x =v0xt+12 axt

2 → x=v0xtvx =v0x + axt→ vx =v0x

Vertical motion: Acceleration due to gravity ==> ay= ±g

- usual equations for constant acceleration

3) Projectile Motion (no friction)

Example: Falling care package

x

Find x.

Step 1: Find t from vertical motion

Step 2: Find x from horizontal motion

3) Projectile Motion (no friction)

Example: Falling care package

x

Step 1:

Given ay, y, v0y

Solve

y =v0yt+12 ayt

2 =−12 gt2

t=2yg

=14.6 s

3) Projectile Motion (no friction)

Example: Falling care package

x

Step 2:

x =v0xt=1680 m

3) Projectile Motion (no friction)b) Nature of the motion: What is y(x)?

Eliminate t from y(t) and x(t):

x =v0xt ; y=v0yt−12 gt2

t=x

v0x

x

y

Substitute:

y =v0y

v0x

⎛

⎝⎜⎞

⎠⎟x−

g2v0x

2

⎛

⎝⎜⎞

⎠⎟x2

y(x) is a parabola

3) Projectile Motion (no friction)

c) Cannonball physics

Initial velocity rv is given:

Either (v0 x ,v0 y ) or (v0 ,θ)

Find (i) height, (ii) time-of-flight, (iii) range

3) Projectile Motion (no friction)

c) Cannonball physics

Initial velocity rv is given:

Either (v0 x ,v0 y ) or (v0 ,θ)

(i) Height: Consider only y-motion: v0y given, ay=-g knownThird quantity from condition for max height: vy=0

Use v2 =v0y2 + 2ayy

When y=H ,v=0, so

0 =v0y2 −2gH → H =

v0y2

2g

Use y =v0yt+12 ayt

2

When y=0,

0 =v0yt−12 gt2 (Trivial solution: t=0)

Non-trivial solution: t=2v0y

g

3) Projectile Motion (no friction)

c) Cannonball physics

Initial velocity rv is given:

Either (v0 x ,v0 y ) or (v0 ,θ)

(ii) Time-of-flight: Consider only y-motion: v0y given, ay=-g knownThird quantity from condition for end of flight; y=0

For y =0, x=R and t=2v0y

g, so

from x=v0xt, R=2v0xv0y

g

3) Projectile Motion (no friction)

c) Cannonball physics

Initial velocity rv is given:

Either (v0 x ,v0 y ) or (v0 ,θ)

(iii) Range: Consider x-motion using time-of-flight: x=v0xt

or, R =v0

2 (2cosθ sinθ)g

=v0

2 sin(2θ)g