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motion in 2 dimension.pptxxxxxxxxxxxxxxxxxxxxxxx

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Kinematics
Motion in two and three dimensions

All previous motion has been
expressed in one dimension
Our position functions are ��
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
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) = ?axt2
+ vxt + dx
y(t) = ?ayt2
+ vyt + dy
z(t) = ?azt2
+ vzt + dz

y
x
z
��
What is the position function for each dimension
��
Divide each vector, (d, v and a) into it��s component for
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)
y
x
z
d =
v =
a =

Position functions
dx(t) =
dy(t) =
15t
-
1
/2gt
2
+ 10
y
x
z
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) = 3t
2
+ 5
y(t) = -t
2
+ 3t �C 2
z(t) = 2t + 1

x(t) = 3t
2
+ 5
y(t) = -t
2
+ 3t �C 2
z(t) = 2t + 1
x(2) = 17
y(2) = 0
z(2) = 5
vx(t) = 6t
vy(t) = -2t + 3
vz(t) = 2
vx(2) = 12
vy(2) = -1
vz(2) = 2
ax(t) = 6
ay(t) = -2
az(t) = 0
ax(2) = 6
ay(2) = -2
az(2) = 0
IaI=
(62
+(-2)2
)
40 = 2 10

Alternatively ��
x(t) = 3t
2
+ 5
y(t) = -t
2
+ 3t �C 2
z(t) = 2t + 1
d(t) = (3,-1,0)t
2
+ (0,3,2)t + (5,-2,1)
v(t) = 2(3,-1,0)t + (0,3,2)

d(t) = (3,-1,0)t
2
+ (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)2
2
+ (0,3,2)2 + (5,-2,1)
v(2) = 2(3,-1,0)2 + (0,3,2)
a(2) = (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) =
v.cos?
v.sin?
ay(t) =
vy(t) =
dy(t) =
0
v.cos?
0
-g
v.sin?
0

So our position functions
for x and y:
v.cos(??t
-
1
/2gt
2
+ v.sin(??t �� or ��
x(t) =
y(t) =
d(t) =

Write a vector position function for
the situation below:
y
x
z
d(t) = 1
/2(0,-g,0)t2
+(v.cos(?),0,0)t + (0,h,0)

Problem
distance

Position function:
y(t) = -
1
/2gt
2
+ v.sin(??t
Distance travelled occurs when height (y)=0
0 = -
1
/2gt
2
+ v.sin(??t
The time (t) when height (y)=0:
t1(-
1
/2gt2 + v.sin(??) = 0 t1= 0
1

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
= v
2
.2cos(?)sin(??

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) �C 0.5gt) = 0
u.sin(20) �C 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

More Related Content

motion in 2 dimension.pptxxxxxxxxxxxxxxxxxxxxxxx

1. Kinematics Motion in two and three dimensions

2. All previous motion has been expressed in one dimension Our position functions are �� x y

3. Now we can express motion in two and three dimensions x y z

4. 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 These can be grouped in two ways ��

5. 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)

6. 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) = ?axt2 + vxt + dx y(t) = ?ayt2 + vyt + dy z(t) = ?azt2 + vzt + dz

7. y x z �� What is the position function for each dimension �� Divide each vector, (d, v and a) into it��s component for A ball is thrown horizontally from a height of 10 metres and at a velocity of 15 metres per second

8. Vector components (0, 0, 0) (15, 0, 0) y x z d = v = a =

9. Position functions dx(t) = dy(t) = 15t - 1 /2gt 2 + 10 y x z d(t) = 1 /2at2 + vt + c Our position function is:

10. 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) = 3t 2 + 5 y(t) = -t 2 + 3t �C 2 z(t) = 2t + 1

11. x(t) = 3t 2 + 5 y(t) = -t 2 + 3t �C 2 z(t) = 2t + 1 x(2) = 17 y(2) = 0 z(2) = 5 vx(t) = 6t vy(t) = -2t + 3 vz(t) = 2 vx(2) = 12 vy(2) = -1 vz(2) = 2 ax(t) = 6 ay(t) = -2 az(t) = 0 ax(2) = 6 ay(2) = -2 az(2) = 0 IaI= (62 +(-2)2 ) 40 = 2 10

12. Alternatively �� x(t) = 3t 2 + 5 y(t) = -t 2 + 3t �C 2 z(t) = 2t + 1 d(t) = (3,-1,0)t 2 + (0,3,2)t + (5,-2,1) v(t) = 2(3,-1,0)t + (0,3,2)

13. d(t) = (3,-1,0)t 2 + (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)2 2 + (0,3,2)2 + (5,-2,1) v(2) = 2(3,-1,0)2 + (0,3,2) a(2) = (6,-2,0)

14. 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.

15. ax(t) = vx(t) = v.cos? v.sin? ay(t) = vy(t) = dy(t) = 0 v.cos? 0 -g v.sin? 0

16. So our position functions for x and y: v.cos(??t - 1 /2gt 2 + v.sin(??t �� or �� x(t) = y(t) = d(t) =

17. Write a vector position function for the situation below: y x z d(t) = 1 /2(0,-g,0)t2 +(v.cos(?),0,0)t + (0,h,0)

18. Problem distance

19. Position function: y(t) = - 1 /2gt 2 + v.sin(??t Distance travelled occurs when height (y)=0 0 = - 1 /2gt 2 + v.sin(??t The time (t) when height (y)=0: t1(- 1 /2gt2 + v.sin(??) = 0 t1= 0 1

20. 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 = v 2 .2cos(?)sin(??

21. 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?

22. dy(t) = -0.5gt^2 + u.sin(20)t 0 = -0.5gt^2 + u.sin(20)t t(u.sin(20) �C 0.5gt) = 0 u.sin(20) �C 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