2c. motion in 2 and 3 dimensions
TRANSCRIPT
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