Fundamentals of Physics(14)

dv ቤተ መጻሕፍቲ ባይዱdt
dx vdt
Indefinite integral
Indefinite integral
0
dv adt a dt dx vdt (v
v at C
To find constant C, let
1 2 2
at )dt
x v0t at C
2-6 Constant Acceleration: A Special Case
v v0 at
(2-11)
Eliminate t :
x x0 vavg t
(2-12)
v v 0 2a( x x0 ) (2-16)
2 2
(2-13) vavg 1 1 2 (v0 v) x x0 2 (v0 v)t (2-17)
Eliminate a :
vavg v0 at
1 2
(2-14)
2
Eliminate v0 :
1 2
x x0 vt at
2
(2-18)
x x0 v0t at
1 2
(2-15)
2-7 Another Look at Constant Acceleration
Eq. 2-11 and 2-15 can be ontained by integration
(2-7)
★ Instantaneous acceleration
v dv d dx d x a lim 2 t 0 t dt dt dt dt
2
(2-8) (2-9)
2-6 Constant Acceleration: A Special Case ★ In many cases the The average velocity 1 acceleration is (2-13) v ( v v ) avg 0 2 constant v v0 ● a aavg Substituting 2-11 for v :
with constant acceleration. Eliminate the air resistance, toss an object either up or down, the object accelerates downward at a certain constant rate: free-fall acceleration g At sea level in Earth’s midlatitudes, The value of g varies slightly with latitude and with elevation. *** Note: For free fall *** (1) The directions of motion are now along a verticle axis with the positive direction of y upward. (2) The free-fall acceleration, being downward, is 2 now negative!
(2-11)
x x0 v0t at
1 2
2 2
2
y y0 v0t gt
1 2
2
(2-15)
v v 0 2a( x x0 ) (2-16) v v 0 2g ( y y0 )
2 2
(2-16)
P24. Sample Problem 2-7
(2-15)
(2-12)
Eq. 2-11 and 2-15: basic equations for constant acceleration. They can be combined in three ways to Yield additional equations.
2-6 Constant Acceleration: A Special Case
●
x
vavg
x2 x1 x t t2 t1
(2-2)
Run 400m
★ Average Speed:
average speed
●

total distance time interval
In 50s
(2-3)
t
vavg savg ?
2-4 Instantaneous Velocity and Speed
Motion Along a Straight Line
Motion Position and Displacement Average Velocity and average Speed Instantaneous Velocity and Speed Acceleration Constant Acceleration: A Special Case Another Look at Constant Acceleration Free-Fall Acceleration
★ In many cases the acceleration is constant v v0 ● aa
avg
The average velocity
vavg (v0 v)
1 2
(2-13)
Substituting 2-11 for
Recast this equation:
t 0
2-5 Acceleration
★ Average Acceleration When a particle’s velocity changes, the particle is said to undergo acceleration. Average Acceleration:
aavg
v2 v1 v t2 t1 t
●
negative direction
-3
-2
-1
0
1
2
3
x ( m)
Origin
Displacement vector
●
magnitude：the distance between the original and final positions. ★ Displacement direction: from an original position to final position. x x2 x1 Can be represented by a (3m) (3m) 6m plus sign or a minus sign.
x x0 t0 C x0
1 2
v v0 at
t 0 v v0 C v0
To find constant C’, let
(4-11)
x x0 v0t at
2
(4-15)
2-8 Free-Fall Acceleration
★ An important example of straight-line motiom
a g 9.8m / s
2-8 Free-Fall Acceleration
Horizontal motion
( constant acceleration )
Vertical motion
( free fall )
v v0 at
(2-11)
v v0 gt
(2-15)
2-1 Motion
★ The motion is along a straight line only (in one dimension).
★The moving object is either a particle or an object that moves like a particle. ★ Forces will not be discussed.
★ a reference frame (参考系) must be chosen and a coordinate system (坐标系) must be constructed on it.
2-2 Position and Displacement ★ Position in x axis Negative Positive reference point: Origin
t 0
(2-11)
Recast this equation:
vavg v0 1 2 at
1 2
(2-14)
2
v v0 at
●
Substituting 2-14 into 2-12:
Similarly
vavg
x x0 vavg t
x x0 t 0
x x0 v0t at
★
Instantaneous Velocity
x dx v lim t 0 t dt
●
(2-4)
t 0 x 0, vavg v
As
at a given instant: (1) v is the derivative of x with respect to t (P16 sample problem 2-3). (2) v is the slope of the position-time curve at the point at that instant (P16. sample problem 2-2). ★ Speed is the magnitude of velocity.
vavg v0 1 2 at
v:
(2-14)
v v0 at
vavg
(2-11)
Substituting 2-14 into 2-12: