复旦大学：《大学物理》课程教学资源（PPT课件，英文）Chapter 2 Motion in one dimension

Chapter 2 Motion in one dimension Kinematics Dynamics

Chapter 2 Motion in one dimension Kinematics Dynamics

Section2 -3 Position, velocity and acceleration vectors 1。 Position vector At any particular time t, the Z particle can be located by its X, y and z coordinates, which are the three components of the posItion vector k F=xi+y计+zk y X where i, and k are the Fig 2-11 cartesian unit vectors

At any particular time t, the particle can be located by its x, y and z coordinates, which are the three components of the position vector : where , and are the cartesian unit vectors. Section2-3 Position, velocity and acceleration vectors 1. Position vector  k  j  i r →    r = x i+ y j+ z k  x y z Fig 2-11  i  j  k r  O

2 Displacement(位移) We defined the displacement y t vector Ar as the change in position vector from t, to t2 t △F=- O x Fig 2-12 Note: 1) Displacement is not the same as the distance traveled by the particle 2)The displacement is determined only by the starting and ending points of the interval

We defined the displacement vector as the change in position vector from t1 to t2 . 2. Displacement (位移) 2 1 r r r     = − r   Note: 1) Displacement is not the same as the distance traveled by the particle. 2) The displacement is determined only by the starting and ending points of the interval. y z 1 r  1 t 2 t= t t= x Fig 2-12 2 r  s O r  

扩(=x+y1j+1k =xi+vitek Then the displacement is =(x2-x1)i+(y2-y1)j+(z2-21k Direction: from start point to end point Magnitude: A-=VAx+Ay2+A22

2 2 2 Magnitude : r = x +y +z  Direction: from start point to end point if r x i y j z k     1 = 1 + 1 + 1 r x i y j z k     2 = 2 + 2 + 2 2 1 r r r     = − Then the displacement is x x i y y j z z k    ( ) ( ) ( ) = 2 − 1 + 2 − 1 + 2 − 1 

The relationship between△fand△s In general △r≠△S △s Can △F=△s Yes. for two cases 1)1D motion without 1ox changing direction 2)When△t→>0 after take limit: dr=ds

The relationship between and : r   s 1 r  P1 2 r  P2 r   x y O z    r s s In general, Can ? r = s  Yes, for two cases: 1) 1D motion without changing direction 2) When after take limit: t → 0, dr = ds 

The difference between|△;and△r(△E (A≡△元) △: magnitude of△F Ar: the change of length B/△ of position vectors △ x2+y2+ Note △F≠△r

The difference between and ( ): r  r  r   r 1 r  P1 2 r  r P2   x y O r z 2 1 2 1 2 1 − x + y + z 2 2 2 2 2 2 r = x + y +z Note r : magnitude of   r   : the change of length of position vectors r ( r r )     r  

When At->0. after take limit △F→)cF △产|GF △s→>C=F ∧y→d △→>=h≠l

When after take limit: t → 0, dr dr    r dr    → | r | | dr |    → s →ds r →dr r d r    → | dr |  =

3. velocity and speed a. The average velocity in any interval is defined to be displacement divided by the time interval △F △t (27) when we use the term velocity we mean the instantaneous velocity b. To find the instantaneous velocity, we reduce the size of the time interval At that is M→>0 and then△r→>0. v(O)=1inF(t+△)-F() lim t→)0 △t->0 △tdt 2-9)

3.velocity and speed a.The average velocity in any interval is defined to be displacement divided by the time interval, (2-7) when we use the term velocity, we mean the instantaneous velocity. b. To find the instantaneous velocity, we reduce the size of the time interval , that is t → 0 and then .  → 0 → r dt dr t r t r t t r t v t t t      =   =  +  − =  →  → lim lim 0 0 ( ) ( ) ( ) t r va v   =   t (2-9)

In cartesian coordinates dr= dx i+dy j+ dzk dr dx dv dz +=j+,k dtdt dt The vector v can also be written in terms of its components as: v=vi+vi+vk dy Vx dt vy dt v: dt (2-12)

   = = + + k dt dz j dt dy i dt dx dt dr v   The vector can also be written in terms of its components as: (2-11) v  v v i v j v k x y z     = + + dt dx vx = dt dy vy = dt dz vz = (2-12)    dr = dx i+ dy j+ dz k  In cartesian coordinates:

Discussion The position vector of a moving particle at a moment is r(x, y The magnitude of the velocity of the particle at the moment is: dr (A) dt √(B) dt (C) (D) dt v dt

Discussion The position vector of a moving particle at a moment is . The magnitude of the velocity of the particle at the moment is: r(x, y)  t r d d t r d | d |  （A） （B） t r d d  2 2 ) d d ) ( d d ( t y t x （C） （D） + √ √