Tutorial the problem parts plus the hints in

2. Tutorial (the problem parts plus the hints in MasteringPhysics walk you through how to do it)
3. Problem 1.02
(a) The motion diagrams are shown below.
(b) In the reference frame of runner 1, he sees runner 2 coming toward him at a speed equal to
v
1
+ v
2
= 2 v .
4. Problem 1.11
With the conversion factors for distance and time
1mi = 5280 ft = 1 . 609 km = 1609 m
1 h = 60 min = 3600 s
we obtain the following conversion factors for speed:

1mph = 1
mi
h
⋅
1h
3600s
⎛
⎝
⎜
⎞
⎠
⎟
⋅
1609m
1mi
⎛
⎝
⎜
⎞
⎠
⎟
= 0 . 447m/s
1mph = 1
mi
h
⋅
1 . 609km
1mi
⎛
⎝
⎜
⎞
⎠
⎟
= 1 . 609km/h
1km/h = 1
km
h
⋅
1000m
1km
⎛
⎝
⎜
⎞
⎠
⎟
⋅
1h
3600s
⎛
⎝
⎜
⎞
⎠
⎟
= 0 . 278m/s
(a)
36 mph = 36 mph
( )
⋅
1 . 609 km/h
1mph
⎛
⎝
⎜
⎞
⎠
⎟
= 57 . 9 km/h = 36 mph
( )
⋅
0 . 447 m/s
1mph
⎛
⎝
⎜
⎞
⎠
⎟
= 16 . 1m/s
(b) 349 km/h = 349 km/h
( )
⋅
1mph
1 . 609 km/h
⎛
⎝
⎜
⎞
⎠
⎟
= 217 mph = 349 km/h
( )
⋅
0 . 278 m/s
1km/h
⎛
⎝
⎜
⎞
⎠
⎟
= 96 . 9m/s
(c) 980 m/s = 980 m/s
( )
⋅
1km/h
0 . 278 m/s
⎛
⎝
⎜
⎞
⎠
⎟
= 3528 km/h = 3528 km/h
( )
⋅
1mph
1 . 609 km/h
⎛
⎝
⎜
⎞
⎠
⎟
= 2192mph
5. Problem 1.25
Sketch and translate Both Gabriele and Xena, the objects of interest, are riding their bicycles at
constant speeds, but in opposite directions. We choose a reference frame with Earth as the object of
reference.
Simplify and diagram We assume that Gabriele and Xena are point-­‐like objects. The origin of the
coordinate system will be the zero mark on the bike path, where an observer stands, and the positive
direction will be toward east.
Represent mathematically The positions of Gabriele and Xena with respect to the zero mark of the
bike path as a function of time are given by
x
G
t
( )
= 3 . 0 × 10
3
m − 8 . 0 m/s
( )
t
x
X
t
( )
= 1 . 0 × 10 3 m + 6 . 0 m/s
( )
t
Solve and evaluate
(a) The functions describing the positions of Gabriele and Xena with respect to Earth are: