1.5 Frames of Reference and
Relative Velocity
A frame of reference is a coordinate
system relative to the motion of an
object.
Keyword: RELATIVE
Relative Velocity: the velocity relative
to a specific frame of reference
Example #1:
Consider a person in a train. The train is
moving 30.0 m/s [S].
a) If the person is sitting on the train, what
is the person’s velocity relative to the
train?
b) If the person is sitting on the train, what
is the person’s velocity relative to the
ground?
c) If the person is walking on the train
at 1.0 m/s [S], what is the person’s
velocity relative to the ground?
d) If the person is walking on the train
at 1.0 m/s [N], what is the person’s
velocity relative to the ground?
e) If the person is running on the train
at 8.0 m/s [W], what is the person’s
velocity relative to the ground?
Example #1 Solution
a) The person’s velocity relative to
the train is zero, since sitting, thus
not moving relative to the train.
→
b) vpg = velocity of the person relative to
the ground
→
vpt = velocity of the person relative to
the train
→
vtg = velocity of the train relative to the
ground
→
→
→
vpg = vpt + vtg
= 0 m/s +(–30.0 m/s)
= – 30.0 m/s
= 30.0 m/s [S]
The person’s velocity relative to the
ground is 30.0 m/s [S].
→
→ →
c) vpg = vpt + vtg
= –1.0 m/s +(–30.0 m/s)
= – 31.0 m/s
= 31.0 m/s [S]
The person’s velocity relative to the
ground is 31.0 m/s [S].
→
d) vpg
→
→
= vpt + vtg
= +1.0 m/s +(–30.0 m/s)
= – 29.0 m/s
= 29.0 m/s [S]
The person’s velocity relative to the
ground is 29.0 m/s [S].
e) Since this question is 2D, the person’s
velocity relative to the ground can NOT
be calculated simply by adding or
subtracting the numbers. The solution is
to use the components method or using
the Sine Law and Cosine Law.
→
→
→
vpg
= vpt + vtg
vpg2 = vpt2 + vtg2
2
vpg = (8.0
2
vpg = 64
2
m/s)
2
2
m /s
+ 900
vpg2 = 964 m2/s2
vpg = 31.0 m/s
+ (30.0
2
m/s)
2
2
m /s
tan θ = o/a
tan θ = vpt
vtg
tan θ = 8.0 m/s
300 m/s
tan θ
= 0.2667
θ
= tan-1(0.2667)
= 14.9º
Thus the direct is [S 14.9º W].
There the person’s velocity relative to
the ground is
31.0 m/s [S 14.9º
W].
Example # 2:
A plane is flying due west at a speed of
200 km/h into a 50 km/h wind
blowing towards the east. What is the
plane’s velocity relative to the ground?
Solution
→
vpg = velocity of the plane relative to the
ground
→
vpa = velocity of the plane relative to the air
→
vag = velocity of the air relative to the
ground (wind velocity)
→
→
→
vpg
= vpa + vag
→
→
→
vpg = vpa + vag
= (–200 km/h)+50 km/h
= – 150 km/h
= 150 km/h [W]
The plane’s velocity relative to the
ground is 150 km/h [W].
Example # 3:
A plane is flying at 200 km/h [N 40
º E] relative to the air. There is a
wind blowing due east at 50km/h.
What is the plane’s ground velocity?
Solution
vpg2 = vpa2 + vag2 – 2vpavagcos
= (200 km/h)2 + (50 km/h)2
– 2(200 km/h)(50 km/h) cos(40º + 90º)
= 40 000 km2/h2 + 2500 km2/h2
+ 12856 km2/h2
= 55356 km2/h2
vpg = 235.28 km/h
sin A
a
sin θ
vag
sin θ
= sin B
b
= sin 130º
vpg
= vag (sin 130º)
vpg
sin θ = (50 km/h) (sin 130º)
235.28 km/h
sin θ
= (50 km/h) (sin 130º)
235.28 km/h
= 0.1628
θ = sin-1(0.1628)
= 9.4º
 = 40º + 9.4º
= 49.4º
Therefore the plane’s ground
velocity is
2.4 X 102
km/h [N 49.4º E].
Example # 4:
From Town A to Town B are on a
straight river. Town B is a 150 km
down stream of Town A. A boat
travelling with its maximum speed can
travel from “A to B” in 5.0 h. The
return trip at maximum speed takes
7.5 h.
a) What is the speed of the current?
b) What is the maximum speed of the
boat?
c) How long would it take to drift
from Town A to Town B?
Solution
→
vbg = velocity of the boat relative to the
ground
→
vbw = velocity of the boat relative to the
water
→
vbg = velocity of the boat relative to the
ground
→
→
→
vbg = vbw + vwg
a)
v = Δd
Δt
(since constant velocity)
Boat travelling from “A to B”
vbg(AtoB) = 150 km
5.0 h
= 30 km/h
→
→
→
vbg(AtoB) = vbw + vwg
Let downstream be “+” and upstream
be “–”
vbg(AtoB)= + vbw + vwg
{Equation #1}
30 km/h
= + vbw + vwg
Boat travelling from “B to A”
vbg(BtoA) = 150 km
7.5h
= 20 km/h
→
→
→
vbg(BtoA)= vbw + vwg
Let downstream be “+” and
upstream be “–”
–vbg(BtoA) = –vbw +vwg
{Equation #2}
–20 km/h = –vbw +vwg
{Equation #1} + {Equation #2}
30 km/h–20 km/h = +vbw +vwg –vbw +vwg
10 km/h
= 2vwg
vwg
= 10 km/h
2
vwg
= 5.0 km/h
Therefore the speed of the current is 5.0
km/h.
b) {Equation #1} – {Equation #2}
30 km/h –(–20 km/h)
= + vbw + vwg –(– vbw+ vwg)
30 km/h +20 km/h
= + vbw + vwg + vbw – vwg
50 km/h
= 2vbw
2vbw = 50 km/h
vbw = 50 km/h
2
2vbw = 50 km/h
vbw = 50 km/h
2
vbw = 25 km/h
Therefore the maximum speed
of the boat is 25 km/h.
c) The boat drifting from Town A
to Town B
v = Δd
Δt
vΔt
= Δd
Δt = Δd
v
Δt = Δd
v
= 150 km
5.0 km/h
= 30 h
Therefore it would take the boat
30 h to drift from Town A to Town B.
Practice Question #1:
A plane is traveling due North with an
airspeed of 200 km/h. There is a 50
km/h wind blowing due east. What is
the velocity of the plane relative to the
ground?
Practice Question #2:
A plane wants to head due North,
unfortunately there is a wind blowing due
west at 50 km/h. The plane has an air
speed of 200 km/h. What should the
direction of the plane be flying relative to
the air? What is the plane’s velocity
relative to the ground?
1.5 Practice Questions
From Nelson 12 Physics (pdf text)
Page 56 Questions 1-5
Page 57 Questions 5-6
(pictures of question on next pg)