motion in 2 dimension.pptxxxxxxxxxxxxxxxxxxxxxxx

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1. Kinematics Motion in two and three dimensions

2. All previous motion has been expressed in one dimension Our position functions are … d(t) = c = vt + c = 1 /2at 2 + vt + c 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 d x d y d z v v x v y v z a a x a y a z These can be grouped in two ways …

5. x y z d d x d y d z v v x v y v z a a x a y a z Each displacement, velocity and acceleration vector can be divided into their x, y and z components … d = (d x , d y , d z ) v = (v x , v y , v z ) a = (a x , a y , a z )

6. x y z d d x d y d z v v x v y v z a a x a y a z …or, a position function can be found for each dimension: x, y and z. x(t) = ½a x t 2 + v x t + d x y(t) = ½a y t 2 + v y t + d y z(t) = ½a z t 2 + v z t + d z

7. y x z 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

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

9. Position functions dx(t) = dy(t) = dz(t) = 15t - 1 /2gt 2 + 10 y x z d(t) = 1 / 2 at 2 + 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 – 2 z(t) = 2t + 1 Find the magnitude of the acceleration, velocity and position vectors when t=2.

11. x(t) = 3t 2 + 5 y(t) = -t 2 + 3t – 2 z(t) = 2t + 1 x(2) = 17 y(2) = 0 z(2) = 5 IdI= (17 2 +5 2 ) 314 vx(t) = 6t vy(t) = -2t + 3 vz(t) = 2 vx(2) = 12 vy(2) = -1 vz(2) = 2 IvI= (12 2 +(-1) 2 + 2 2 ) 149 a x (t) = 6 a y (t) = -2 a z (t) = 0 a x (2) = 6 a y (2) = -2 a z (2) = 0 IaI= (6 2 +(-2) 2 ) 40 = 2 10

12. Alternatively … x(t) = 3t 2 + 5 y(t) = -t 2 + 3t – 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) a (t) = 2(3,-1,0) = (6,-2,0)

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) 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)

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) = dx(t) = v.cos  v.sin  a y (t) = v y (t) = d y (t) = v.cos  -g v.sin 

16. So our position functions for 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) =

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

18. Problem How far will the ball travel if kicked at velocity ‘v’ and at angle ‘  ’? 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 /2gt2 + v.sin(  = 0 t2 = - v.sin(  = t2= 2v.sin(  - 1 /2g g

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(  g = v 2 .sin(2  g

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) – 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