1. Physics 211 Space - time & space-space diagrams Kinetic Equations of Motion Projectile motion Uniform circular motion Moving coordinate systems Relative motion Galilean Transformation of coordinates 3: Two Dimensional Motion

2. r(t 1 ) r(t 2 ) x y space-space diagram

4. at 1   dvt 1  dt  d 2 xt 1  2 i  d 2 yt 1  2 j at 1   dvt 1   a x t 1  i  a y t 1  j speed= vt   vt   dxt  dt     2  dyt  dt     2 acceleration = at   at   d 2 xt  dt 2     2  d 2 yt  2     2 average quantities= final value-initial value time taken

5. acceleration due force of gravity near the earths surface is approximately constant Neglect air resistance Neglect rotation of earth Then we can use kinetic equations of motion for projectile motion

6. i j y x

7. a  g  9.81j m s 2 r0   0 rt   1 2 gt 2     j  v x 0  t  i  v y 0  t  j  v x 0  t  i  v y 0  t  1 2 2     j  xt   v x 0  t yt   v y 0  t  1 2 2     y x Horizontal and vertical positions

8. velocity in x direction x t  = dx dt  v x 0  velocity in y direction y t  = dy dt  v y 0   9.81t when projectile reaches highest point v y (t)=0  v y 0   9.81t  0  t high  v y 0  9.81  xt high   v x 0  v y 0  9.81 ; yt high   v y 0  2 9.81 - 1 2 9. v y 0  9.       2 = 1 2 v y 0  2 9. = v y 0  2 19.62 projectile hits ground when yt   0  t v y 0   1 2 9.81t      0  t  0 or t  2 vy0vy0  9.81  x max  2 v x 0  v y 0  9.81 v v

9. trajectory angle tan   slope of tangent to path at t  0  slope of velocity vector at t  0  v y 0  v x 0   dy dt dx dt  dy dx  tan  1 v y 0  v x 0        y x 

10. initial speed= v0   v x 0  2  v y 0  2 v x 0   v0  cos  v y 0   v0  sin   xt high   v0  2 sin  cos  g  v0  2 sin2  2g yt high   v0  2 sin 2  2g and x range  2 v x 0  v y 0  9.81  2v0  2 sin  cos  g  v0  2 sin2  g Maximun height and range can be expressed in terms of v0  and 

11. y x  v(0) v0  2 sin 2  2g v0  2 sin (2  g

12. Uniform circular motion  r v vt   v=constant; rt   r  position vector rt  = xt ,yt    rcos  t ,rsin  t   rcos  t ,rsin  t    rcos  t,rsin  t   t   t  d  t  dt  constant        is angular speed angular acceleration d  t  dt  0 distance travelled s = r  r  t linear speed=v= ds dt  r 

13. rt   rcos  t i  rsin  t j  r ˆ rt  vt   drt  dt  r  sin  t i  r  cos  t j  r  sin  t i  cos  t j   r  ˆ vt  at   dvt  dt  r  2 cos  t i  2 sin  t j   r  2 cos  t i  sin  t j   r  2 ˆ rt  at   r  2 ˆ rt   v 2 r ˆ rt  at   v 2 r  constant

14. Relative motion observer 1 observer 2 u(t)=u=constant r 2 (t) r 1 (t) r(t)=ut

15. r 1 t   position vector of object in coordinate system r 2 t   r 1 t   r 2 t   ut  r 2 t   r 1 t   ut v 2 t   v 1 t   u a 2 t   a 1 t      Galilean Transformation 1  2 1 2

16. Nonuniform curvilinear motion

17. rt   rt  cos  t  i  sin  t  j  vt   drt  dt  v t  ˆ vt  find unit vector, ˆ ct  perpendicular to ˆ vt  points to center of curvature of path at this point   ˆ vt   ˆ vt   1  d dt ˆ vt   ˆ vt    0  d ˆ vt   ˆ vt   ˆ vt   d ˆ vt   0  ˆ vt   d ˆ vt   0 thus ˆ ct   d ˆ vt  dt d ˆ vt 

18. a t   dvt   d v t  ˆ vt   v t  d ˆ vt  at   a t t  ˆ vt  +a r t  ˆ ct  a r t   v t  2 rt   radial(centripetal) acceleration a t t   rt   t   tangential acceleration rt   distance to center of curvature  