Calculus and Vectors – How to get an A+
7.2 Velocity
A Velocity
Velocity is a vector and the measurement unit is
m / s or km / h .
B Relative Velocity
r
The relative velocity of the object B traveling at v B
r
relative to the object A traveling at v A is given by:
r
r
r
v BA = v B − v A
r
Ex 2. A car is traveling at vc = 100km / h[ E ] , a
r
motorcycle is traveling at vm = 80km / h[W ] , a truck
r
is traveling at vt = 120km / h[ N ] and an SUV is
r
traveling at v s = 100km / h[ SW ] . Find the relative
velocity of the car relative to:
a) motorcycle
b) truck
c) SUV
Ex 1. Convert 5m / s into km / h .
1km
m
1km 3600
5 = 5 1000 = 5
= 5(3.6)km / h = 18km / h
1h
s
1000 1h
3600
r
r
Note. If A is at rest ( v A = 0 ) then:
r
r
v BA = v B
Let first change vectors into algebraic vectors:
r
vc = 100km / h[ E ] = (100,0)km / h
r
vm = 80km / h[W ] = (−80,0)km / h
r
vt = 120km / h[ N ] = (0,120)km / h
r
v s = 100km / h[ SW ] = (−100 cos 45°,−100 sin 45°)
= (−50 2 ,−50 2 )km / h
r
r r
a) vcm = vc − vm = (100,0) − (−80,0) = (180,0)km / h
r
r r
b) vct = vc − vt = (100,0) − (0,120) = (100,−120)km / h
c)
C Boat Velocity
The boat velocity relative to ground is vector sum
r
r r
vcs = vc − v s = (100,0) − (−50 2 ,−50 2 )
= (100 + 50 2 ,50 2 )km / h
D Plane Velocity
The plane velocity relative to ground is vector sum between
between the boat velocity relative to water vbw and the plane velocity relative to air v pa and the air velocity
the water velocity relative to ground v wg :
relative to ground vag :
vbg = vbw + v wg
v pa = v pa + v ag
Ex 3. A river flows eastward with 4m / s . A
motorboat heads downstream the river between
two towns which are 50km apart along the south
bank of the river. If the motorboat speed in still
water is 12m / s , find:
a) the speed of the motorboat relative to the ground
when traveling downstream
Ex 4. A plane is scheduled to travel from the airport A to an
airport B where AB = 600km[060°] . The speed of the plane
relative to air is 300km / h and a strong wind of 100km / h is
blowing eastward.
a) Draw a diagram to illustrate the situation.
r
v wg = 4m / s[ E ]
r
vbw = 12m / s[ E ]
r
r
r
vbg = vbw + v wg = 12m / s[ E ] + 4m / s[ E ] = 16m / s[ E ]
r
∴ || vbg ||= 16m / s
7.2 Velocity
©2010 Iulia & Teodoru Gugoiu - Page 1 of 2
Calculus and Vectors – How to get an A+
b) the time required to cover the distance between
the towns downstream
D = 50km
D
50km
=
= 3125s = 52.08 min
t= r
|| vbg || 16m / s
b) In what direction should the pilot head the plane?
sin β
sin 30°
sin β sin 30°
= r
⇒
=
⇒
r
|| vag || || v pa ||
100
300
⎛ sin 30° ⎞
⎟ = 9.59°
⎝ 3 ⎠
α = 60° − 9.59° = 50.41°
c) the speed of the motorboat relative to the ground
The pilot should head the plane in direction [ N 50.41° E ]
when traveling upstream
β = sin −1 ⎜
c) What is the speed of the plane relative to ground?
r
v wg = 4m / s[ E ]
r
vbw = −12m / s[ E ]
r
r
r
vbg = vbw + v wg = −12m / s[ E ] + 4m / s[ E ] = −8m / s[ E ]
r
∴|| vbg ||= 8m / s
d) the time required to cover the distance between
the towns upstream
D
50km
=
= 6250 s = 104.17 min
t= r
|| vbg || 8m / s
Ex 5. A river is 800m wide and flows eastward at
10m / s . Peter is driving a motorboat heading
always perpendicular to the current. The speed of
the motorboat in still water is 20m / s .
a) Draw a diagram to illustrate the situation.
r
|| v pg ||
=
r
|| v pa ||
⇒
r
|| v pg ||
sin(180° − 30° − 9.59°) sin 30°
sin 140.41°
r
300 sin 140.41°
|| v pg ||=
≅ 382.37km / h
sin 30°
=
300
sin 30°
d) How long will the trip last?
D
600km
t= r
=
≅ 1.57 h
|| v pg || 382.37 km / h
Ex 6. Jane can swim at 5m / s in still water. She wishes to
swim across a river 200m wide to a point directly opposite
from where she is standing. The river flows westward at
4m / s and she is standing on the South bank of the river.
a) Draw a diagram to illustrate the situation.
b) What is the speed of the boat relative to ground? b) What is the speed of Jane relative to ground?
r
r
r
r
r
r
|| vbg ||= || vbw || 2 + || v wg || 2 = 20 2 + 10 2 = 500
|| v Jg ||= || v Jw || 2 − || v wg || 2 = 5 2 − 4 2 = 3m / s
r
c) In what direction must Jane head?
∴ || vbg ||= 10 5m / s ≅ 22.36m / s
r
|| v wg || 4
c) How long does it take to cross the river?
sin θ = r
= ⇒ θ = sin −1 (4 / 5) = 53.13°
||
||
5
v
D
800m
Jw
t= r
=
= 40 s
Jane must head into direction [ N 53.13°W ] .
|| vbw || 20m / s
d) How long does it take to cross the river?
d) How much downstream does Peter reach the
opposite bank?
D
200m
r
t= r
=
= 66.67 s
AB =|| v wg || t = 10(40)m = 400m
|| v Jg || 3m / s
Reading: Nelson Textbook, Pages 365-368
Homework: Nelson Textbook: Page 369 #1, 3, 4, 6, 9, 11, 13, 14
7.2 Velocity
©2010 Iulia & Teodoru Gugoiu - Page 2 of 2