Chapter 2:
Motion in One Dimension
EXAMPLES
Example 2.1 Displacement


x1 = 30 m x2 = 10 m
Displacement is a VECTOR
x  x f  xi  10m  30m  20m
Example 2.2 Average Velocity & Speed


Suppose the person walks during 50
seconds.
Displacement
x  x f  xi  40m  0m  40m

Distance (d) = 100m
Average velocity:
vaverage  vx 

x 40m

 0.8m / s
t
50 s
Average Speed:
Average speed 
total distance 100m

 2.0m / s
total time
50 s
v average  v x 
Xi
x x f  xi

t
t
Xf
Example 2.3 Instantaneous & average
velocities

If an objects moves at uniform velocity (constant),
then: Instantaneous velocity and average
velocity at any instant (t) are the same.
Instantaneous = average
Example 2.4 Instantaneous & average
velocities


Are Instantaneous velocity and Average
velocity at any instant t the same? NOT
ALWAYS!!!!
Example: A car starts from rest, speed up to
50km/h, remains at that speed for a time. Slow
down to 20 km/hr in a traffic jam, the finally
stops. Traveling a total of 15 km in 30 min (0.5
hr).
Example 2.4, cont
Average velocity 
x 15km

 30.0km / h
t 0.50h
2.4 Acceleration
Example 2.5 Average Acceleration

ax (+), vx(+) Speeding Up!!
ax 

v f  vi
t f  ti

75km / h  0
km
1000m
 15
 15
 4.2m / s 2
5.0s  0
hs
3600s  s
ax (), vx() Speeding Up!!
ax 
v f  vi
t f  ti

 15.0m / s  (5m / s)  10.0m / s

 2.0m / s 2
5.0s  0
5.0s
Example 2.6 Average Acceleration

ax (+), vx() Slowing Down!!
ax 

v f  vi
t f  ti

 5.0m / s  (15.0m / s) 10.0m / s

 2.0m / s 2
5.0s  0
5.0s
ax (), vx(+) Slowing Down!!
ax 
v f  vi
t f  ti

5.0m / s  15.0m / s  10.0m / s

 2.0m / s 2
5.0s  0
5.0s
Example 2.7 Conceptual Question


Velocity and acceleration are both vectors (they have
magnitude & direction).
Are the velocity and the acceleration always in the
same direction?
NO WAY!!
Example 2.8 Conceptual Question


Velocity and acceleration are both vectors (they have
magnitude & direction).
Is it possible for an object to have a zero
acceleration and a non-zero velocity?
YES!!!
Drive 65 miles/h on
the Freeway
Example 2.0 Conceptual Question


Velocity and acceleration are both vectors (they have
magnitude & direction).
Is it possible for an object to have a zero velocity
and a non-zero acceleration?
YES!!!
Start your car!!!
Material for the Midterm

Examples to Read!!!


Example 2.5 (Text book Page 31)
Example 2.8 (Text book Page 37)
2.6 Constant Acceleration
Example 2.10 Free Fall Example

Initial velocity at A is upward (+) and
acceleration is g (– 9.8 m/s2)

At B, the velocity is 0 and the
acceleration is g (– 9.8 m/s2)

At C, the velocity has the same
magnitude as at A, but is in the
opposite direction

The displacement is – 50.0 m
(it
ends up 50.0 m below its starting
point)
Example 2.10, cont

(1) From (A) → (B)
Vyf(B) = vyi(A) + ayt(B)
 0 = 20m/s + (–9.8m/s2)t(B)
t = t (B) = 20/9.8 s = 2.04 s
ymax = y(B) = y(A) + vyi(A)t +
½ayt2
y(B) = 0 + (20m/s)(2.04s) +
½(–9.8m/s2)(2.04s)2
y(B) = 20.4 m
Example 2.10, cont
(2) From (B) → (C): y(C) = 0
y(C) = y(A)+ vyi(A) t – ½ayt2
 0 = 0 + 20.0 t – 4.90t2
(Solving for t): t(20 – 4.9t) = 0
 t = 0 or t(C) = t = 4.08 s

vyf(C) = vyi(A) + ayt (C)
 vyf(C) = 20m/s + (– 9.8m/s2)(4.08 s)

vyf(C) = –20.0 m/s
Example 2.10, cont
(3) From (C) → (D)
Using position (C) as the reference
point the t at (D) position is not 5.00s.

It will be:
t (D) = 5.00 s – 4.08 s = 0.96
 vyf(D) = vyi(C) + ayt (D)
 vyf(D) = -20m/s + (– 9.8m/s2)(0.96 s)
vyf(D) = – 29.0 m/s
 y(D) = y(C) + vyi(C)t + ½ayt2
 y(D) = 0 – (29.0m/s)(0.96s) –
(4.90m/s2)(0.96s)2 = – 22.5 m

y(D) = –22.5 m
Example 2.11 (Problem #66 page 54)




d
1
9.80 t12
2
From the free fall of the rock the distance will be:
From the sound de same distance will be: d  336t 2
But: t1 + t2 = 2.40s  t1 = 2.40 – t2
Replacing (t1 ) into the first equation and equating to the second:
336t 2  4.90 2.40  t 2 
2
t2 

359.5  359.52  4 4.90 28.22 
4.90t 22  359.5t 2  28.22  0

9.80

d  336t 2  26.4 m
.
t2 
359.5  358.75
 0.076 5 s
9.80
Example 2.12 Objective Question #13
A student at top of the building of height h throws one ball upward
with speed vi and then throws a second ball downward with the
same initial speed, vi . How do the final velocities of the balls
compare when they reach the ground?
After Ball 1 reaches
maximum height it falls
BALL
2
BALL 1
back downward
passing the student
with velocity –vi . This
+vi
velocity is the same as
Ball 2 initial velocity, so
- vi
after they fall through
- vi
equal height h, their
impact speeds will
h
h
also be the same!!!

Material for the Midterm

Material from the book to Study!!!



Objective Questions: 2-13-16
Conceptual Questions: 6-7-9
Problems: 3-6-11-16-17-20-29-42-44