Kinematics

Motion in two and three dimensions

All previous motion has been expressed in one dimension

Our position functions are …

d(t) = c

= vt + c

= 1

/2at2

+ vt + c

x

y

Now we can express motion in two and three dimensions

x

y

z

To understand motion in 2/3 dimensions, distance, velocity and acceleration need to be divided into 3 directions: x, y and z

x y z

d dx dy dz

v vx vy vz

a ax ay az

These can be grouped in two ways …

x y z

d dx dy dz

v vx vy vz

a ax ay az

Each displacement, velocity and acceleration vector can be divided into their x, y and z components …

d = (dx , dy , dz)v = (vx , vy , vz)a = (ax , ay , az)

x y z

d dx dy dz

v vx vy vz

a ax ay az

…or, a position function can be found for each dimension: x, y and z.

x(t) = ½axt2 + vxt + dx

y(t) = ½ayt2 + vyt + dy

z(t) = ½azt2 + vzt + dz

y

xz● What is the position function for each dimension● Divide each vector, (d, v and a) into it’s component for

each dimension.

A ball is thrown horizontally from a height of 10 metres and at a velocity of 15 metres per second

Vector components

(0, 0, 0)

(15, 0, 0)

(0, -g, 0)

y

xzd =v =a =

Position functions

dx(t) =

dy(t) =

dz(t) =

15t

-1/2gt

2 + 10

0

y

xz

d(t) = 1/2at2 + vt + c

Our position function is:

Problem

The motion of a creature can be described in 3 dimensions by the following equations for the position in the x, y and z directions:

x(t) = 3t2

+ 5

y(t) = -t2

+ 3t – 2

z(t) = 2t + 1

Find the magnitude of the acceleration, velocity and position vectors when t=2.

x(t) = 3t2

+ 5

y(t) = -t2

+ 3t – 2

z(t) = 2t + 1

x(2) = 17

y(2) = 0

z(2) = 5

IdI=

(172

+52

)

314

vx(t) = 6t

vy(t) = -2t + 3

vz(t) = 2

vx(2) = 12

vy(2) = -1

vz(2) = 2

IvI=

(122

+(-1)2 + 2

2)

149

ax(t) = 6ay(t) = -2az(t) = 0

ax(2) = 6ay(2) = -2az(2) = 0

IaI=(62+(-2)2)

40 = 2 10

Alternatively …

x(t) = 3t2 + 5

y(t) = -t2

+ 3t – 2

z(t) = 2t + 1

d(t) = (3,-1,0)t2

+ (0,3,2)t + (5,-2,1)

v(t) = 2(3,-1,0)t + (0,3,2)

a(t) = 2(3,-1,0) = (6,-2,0)

d(t) = (3,-1,0)t2

+ (0,3,2)t + (5,-2,1)

v(t) = 2(3,-1,0)t + (0,3,2)

a(t) = (6,-2,0)

d(2) = (3,-1,0)22

+ (0,3,2)2 + (5,-2,1)

v(2) = 2(3,-1,0)2 + (0,3,2)

a(2) = (6,-2,0)

d(2) = (12,-4,0) + (0,6,4) + (5,-2,1) = (17,0,5)

v(2) = (12,-4,0) + (0,3,2) = (12,-1,2)

a(2) = (6,-2,0) = (6,-2,0)

Projectile motion

A ball is kicked with a velocity ‘v’ at an angle of ‘’.Write a position function for each the x and z direction.

ax(t) =

vx(t) =

dx(t) =

v.cos

v.sin

ay(t) =

vy(t) =

dy(t) =

0v.cos0

-gv.sin0

So our position functionsfor x and y:

v.cos(t

-1/2gt

2 + v.sin(t … or …

1/2(0,-g,0)t

2 +(v.cos(v.sin(,0)t + (0,0,0)

x(t) =

y(t) =

d(t) =

Write a vector position function for the situation below:

y

xz

d(t) = 1/2(0,-g,0)t2 +(v.cos(),0,0)t + (0,h,0)

Problem

How far will the ball travel if kicked at velocity ‘v’ and at angle ‘’?

distance

Position function:

y(t) = -1/2gt

2 + v.sin(t

Distance travelled occurs when height (y)=0

0 = -1

/2gt2 + v.sin(t

The time (t) when height (y)=0:

t1(-1

/2gt2 + v.sin() = 0 t1= 0

-1/2gt2 + v.sin(= 0

t2 = -v.sin( = t2= 2v.sin(

-1

/2g g

We can now find the distance travelled by …

substituting ‘t2’ into the ‘x’ position function.

x(t) = v.cos(t

x(t2) = v.cos((2v.sin(

g

= v2

.2cos()sin(

g

= v2

.sin(2

g

Problem 2

What is the speed (m/sec) needed for a stunt driver to launch from a 20 degree ramp to land 15 m away? What is his maximum height?

dy(t) = -0.5gt^2 + u.sin(20)t

0 = -0.5gt^2 + u.sin(20)t

t(u.sin(20) – 0.5gt) = 0

u.sin(20) – 0.5gt = 0

0.5gt = u.sin(20)

t = 2u.sin(20)/g

dx(t) = u.cos(20)t

15 = u.cos(20)t

u = 15 / (cos20.t)

= 15

cos20(2u.sin(20)/g)

u^2 = 15g

2.cos20.sin20

u = [15g / (2.cos20.sin20)]^0.5