Transcript Slide 1

Chapter 2
Rectilinear motion:
straight-line kinematics
Kinematics
a description of motion
without considering its cause
Scalars – quantities having only
magnitude (force) Ex. 95 km/hr
Vectors – quantities having both magnitude
and direction
Ex. 95 km/hr N on I35E
Vectors can be added and subtracted via vector
addition, which is generally geometric
however, for straight line motion vector
addition for one direction is algebraic.

When vectors are parallel
(same direction or opposite direction)
Then their sum or resultant is the
algebraic sum or “+” & “-” numbers.

Position – the separation of an object from a

Distance – a measure of separation between

Displacement – the measure of the change
reference point (vector)
two points (a scalar)
in position Δd
(Δd for straight line motion)

All velocities are values with a frame of
reference or a standard relative to which
speeds and/or velocities taken.
70 mi/hr is relative to what frame?
Example
2 m/s
10
m/s
-2 m/s
Vo Observer
on shore
Shore
Vs = 10 m/s = velocity of ship
Vp = velocity of passenger
Vo = velocity relative to observer
of passenger
Vo = ? Vo = Vs + Vp
a) Vo = 10 m/s + 2 m/s = _____m/s
b) Vo = 10 m/s + (-2 m/s) = ______m/s
If the velocity is the same for a given time interval
= Uniform Velocity
d
(m)
Δd
(rise)
Δt
(run)
V = Δd/Δt = slope of the line
= rise/run
Time t
(sec)
#1 Velocity is the slope of a ΔPosition vs time graph

Instantaneous velocity –
velocity at any particular moment in t
d
(m)
t 1t
V at time t = t1
is the slope of the tangent line to the d vs. t curve at t = t1
V = Δd
Δt
Units: m for Δd
v Δt = Δd
*displacements can be obtained from a v vs. t graph.
Δd = area under curve from t1 to t2
(rectangle)
Δd = area under curve from 0 to t1
(triangle)
V
m/s
Δd
Δd
Let t1 = 3 sec &
t2 = 7 sec
0
(t2 –t1)
Δd1 = vΔt = (5m/s)(7sec – 3sec) = (5m/s)(4sec) = 20m
t1
t2
t
sec
Δd2 = vΔt = area of Δ = ½ (3sec – 0sec)(5m/s)
= ½ (3sec)(5m/s)
= 7.5 m
< d = d1 + d2 = 20m + 7.5m = 27.5m displacement>
#2 Change in Displacement is the AREA beneath
a velocity vs. time graph
Acceleration – the rate of change in velocity, a
a = Δv = (v2 – v1)
Δt
(t2– t1)
Units: m/sec = m/sec2
sec
One direction
If a > 0, then velocity is increasing
If a < 0, then velocity is decreasing and is called deceleration
#3 Acceleration is the Slope of a v vs.
Example:
Givens: v2 = 16 m/s v1 = 2 m/s
t2 = 4.5 sec t1 = 1 sec
Equation: a = (v2 – v1) / (t2 – t1)
Plug into equation:
a = (16 m/sec – 2 m/sec) / (4.5 sec – 1 sec)
a = (14 m/sec) / (3.5 sec)
a = _________m/s2
t graph
Change in velocity between t1 & t2
= Δv = area under curve from t1 to t2
a
(m/s2)
Δv
Let t1 = 2 sec & t2 = 6 sec
Δv = a Δt = area of rectangle
0
t1
t2
T (sec)
= (3.6 m/s2)(6 sec – 2 sec)
= (3.6 m/s2)(4 sec)
= ________m/sec
Leads to
#4 The change in velocity is the area
beneath an acceleration vs. time graph.
For uniform acceleration,
a = Δv = vf – vi
Δt
If t1 = 0
tf – ti
Let Δv = vf – vi
vf = final
Solve for vf :
at = vf – vi
Vi + at = Vf
vf = vi + at
Eq. #1
no distance needed
vi = initial
For vf and vi
v = ( vf + vi)
2
( if ti = 0 & di = 0 )
average
v = (Δd) = d
t
t
d = vt = ( vf + vi ) ttotal
“0” eq.
2
No acceleration
Zero initial time and initial distance
Combine Eq. #1 and the “0” Eq.
To create an equation without Vf
Vf= Vi + at
d= (Vf + Vi) t
2
d= ((Vi+ at )+Vi) t
2
d=( 2Vi + at) t
2
d = (Vi + ½ at) t
d = Vit + ½ at2
No Vf
Eq #2
Take Eq. #1 and solve for t:
t= Vf - Vi
a
Substitute that equation for time into Eq. #2
d = Vit + ½ at2
d= Vi(vf-vi) + ½ a(vf-vi)2
a
a
Math MAGIC!!!!!!
d = (Vf2- Vi2)
2a
2ad = Vf2 – Vi2
Vi2 + 2ad = Vf2
No time
Eq. #3
2A pg.44 -1,3,6
2B pg.49 -1,4,5
2C pg.53 -1,3
2D pg.55-1,4
2E pg.58 -3,5
2F pg.64 -1,4
#6 a. 6.4 hr.
b. 77km/hr south
#4 -3.5 x10-3 m/s2
#4 2.5 sec , 32m
#3. a.16m/s b.7sec
5. +2.3 m/s2
#4 a. 3.7m b. 0.76 sec
2A
1. 738 sec
2. 75 sec
3. a)19.8 sec
b) 468 m
7. a) -1.04m/s
b) 0.0 m/s
c) 1.72 m/s
2C
2. t= 120sec
5. x=d= 300m
7. Vi= 209 km/hr
2B 1. Vf= 389 km/hr
2. 23m/s = v
7. a) 50 sec
b) .12 m/s2
8. t= 4.5 sec
2D- 12. vi= 120 m/s
13. a= 1.36 m/s2
14. a=.87m/s
2E- 2. d=1.41 km
5. vi=196 m/s
10. a= -1.51m/s2
2F- 1. vf=-82m/s
6. t=6.82 sec
8. t=3sec