PHYS101
Motion in two dimensions
Spring 2014
Motion in Two Dimensions
1. A motorist drives south at 20.0 m/s for 3.00 min, then turns west and travels at
25.0m/s for 2.00 min, and finally travels northwest at 30.0 m/s for 1.00 min. Let
the positive x axis point east. For this 6.00 min trip, find
(a) the total vector displacement,
(b) the average speed, and
(c) the average velocity.
Solutions:
(a) First we have to calculate the position vectors, then we can calculate the net
displacement:
In the first part of the trip the motorist travels with a constant velocity of
20.0 m/s in negative y direction, therefore the motorist’s position vector
after 3.00 min is:
~r1 = −20.0 m/s · 3.00 min ĵ = −20.0 m/s · 180 s ĵ = −3.60 × 103 m ĵ
m
s
In the second part of the trip the motorist travels for 2.00 min with a constant speed of 25 m/s to the west, i.e. in negative x-direction, as the positive
x direction points to the east. Therefore the displacement for the second
part of the trip :
~r2 = −25.0
m
m
2.00 min î = −25.0 120 s î = −3.00 × 103 m î
s
s
In the third part of the trip the motorist travels with a constant speed of
30.0 ms in northwest direction for 1.00 m. Therefore the displacement for
the third part of the trip is:
m
m
cos(135◦ ) î + 30.0 sin(135◦ ) ĵ)1.00 min =
s
s
m
m
= (−21.2 î + 21.2 ĵ)60.0 s = (−1.27 × 103 m î + 1.27 × 103 m ĵ)
s
s
~r3 = (30.0
So finally the net displacement after 6.00 min is
~r = ~r1 + ~r2 + ~r3 =
= (−3.60 × 103 m ĵ) + (−3.00 × 103 m î) + (−1.27 × 103 m î + 1.27 × 103 m ĵ) =
= (−4.27 × 103 m î − 2.33 × 103 m ĵ) = (−4.27 î − 2.33 ĵ) km
c 2014 Department of Physics, Eastern Mediterranean University
Page 1 of 10
PHYS101
Motion in two dimensions
Spring 2014
The magnitude of the net displacement is then
q
q
2
2
r = r x + ry = (−4.27 km)2 + (−2.33 km)2 = 4.86 km
The direction of the net displacement vector is then
−1 −2.33 km
θ = tan
= 208.6◦
−4.27 km
So the direction is 28.6◦ south of west.
(b) In order to determine the average speed we have to sum up the magnitude
of the displacements and finally divide the sum by the total amount of
time that has passed. Therefore we get for the average speed, denoted by
v av.speed
|~r1 | + |~r1 | + |~r3 |
=
∆t p
3.60 km + 3.00 km + (−1.27 km)2 + (1.27 km)2
=
= 0.0233 km/s = 23.3 m/s
360 s
v av.speed =
(c) We get the magnitude of the average velocity as the ratio of the length of
the displacement vector over the time that has passed.
v av
|~r |
4.86 × 103 m
=
=
= 13.5 m/s
∆t
360 s
The direction of the average velocity vector is in direction of the net displacement vector ~r.
c 2014 Department of Physics, Eastern Mediterranean University
Page 2 of 10
PHYS101
Motion in two dimensions
Spring 2014
2. A fish swimming in a horizontal plane has velocity ~vi = (4.00î + 1.00ĵ)m/s at a
point in the ocean where the position relative to a certain rock is ~ri = (10.0î −
4.00ĵ)m. After the fish swims with constant acceleration for 20.0 s, its velocity is
~v = (20.0î − 5.00ĵ)m/s
(a) What are the components of the acceleration of the fish?
(b) What is the direction of the acceleration with respect to the unit vector î?
(c) If the fish maintains constant acceleration, where is it at t = 25.0s and in
what direction is it moving?
Solutions:
(a)
~v(t) = ~v0 +~at
1
⇒ ~a = (~v(t) − ~v0 ) =
t
1
(20.0î − 5.00ĵ)m/s − (4.00î + 1.00ĵ)m/s =
20.0 s
1
1
((20.0 − 4.00)î + (−5.00 − 1.00)ĵ)m/s =
(16.0î − 6.00ĵ)m/s =
=
20.0 s
20.0 s
= (0.800î − 0.300ĵ)m/s2
~a =
so we get for the components of the acceleration of the fish
a x = 0.800 m/s2 ,
ay = −0.300 m/s2 .
Alternatively we can also calculate the component directly, taking into account that the average acceleration and the instantaneous acceleration are
equal if the acceleration is constant.
20.0 m/s − 4.00 m/s
∆v x
=
= 0.800 m/s2
∆t
20.0 s
∆vy
−5.00 m/s − 1.00 m/s
=
=
= −0.300 m/s2
∆t
20.0 s
ax =
ay
(b) The direction is then
θ = tan
−1
−0.300 m/s2
0.800 m/s2
= −20.6◦ = 339◦
The acceleration vector has the direction 339◦ with respect to the positive
x-axis
c 2014 Department of Physics, Eastern Mediterranean University
Page 3 of 10
PHYS101
Motion in two dimensions
Spring 2014
(c) In order to determine the position we consider the position vector components at t = 25.0 s
1
m
1
m
x (t) = x0 + v x0 t + a x t2 = 10.0 m + 4.00 (25.0 s) +
0.800 2 (25 s)2 = 360 m
2
s
2
s
1 2
y ( t ) = y 0 + v y0 t + a y t
2
1
m
m
−0.300 2 (25 s)2 = −72.7 m
= −4.00 m + 1.00 (25.0 s) +
s
2
s
In order to determine the direction of the motion we have to consider the
final velocity vector at t = 25.0 s. The components of the final velocity
vector at t = 25.0 s are then:
m
m
m
v x (t) = v x0 + a x t = 4.00 + 0.800 2 (25.0 s) = 24.0
s
s
s
m
m
m
vy (t) = vy0 + ay t = 1.00 − 0.300 2 (25.0 s) = −6.50
s
s
s
Then we get for the direction
m
−1 v y
−1 −6.50 s
θ = tan
= tan
= −15.2◦
m
vx
24.0 s
Which corresponds to angle of 165◦ with respect to the positive x-direction.
3. An electron’s position is given by ~r = 3.00tî − 4.00t2 ĵ + 2.00k̂, with t in seconds
and ~r in meters.
(a) In unit-vector notation, what is the electron’s velocity ~v(t)
(b) What is ~v(t)at t = 2.00s in unit-vector notation?
(c) What is the magnitude of ~v?
(d) What is the angle of ~v relative to the positive direction of the x axis?
Solutions:
(a) As the position vector is given as a function of time we can calculate the
velocity easily by
d m
m
d~r (t)
2
~v(t) =
=
3.00tî − 4.00t ĵ + 2.00k̂ = 3.00 î − 8.00 2 tĵ
dt
dt
s
s
(b) Evaluating this result at t = 2s produces
~v(2.00 s) = 3.00
m
m
m
m
î − 8.00 2 (2.00 s)ĵ = 3.00 î − 16.0 ĵ
s
s
s
s
(c) The speed at t = 2.00 s is
r
v=
3.00
m 2 m 2
m
+ −16
= 16.3
s
s
s
c 2014 Department of Physics, Eastern Mediterranean University
Page 4 of 10
PHYS101
Motion in two dimensions
Spring 2014
(d) The angle of ~v relative to the positive direction of the x axis is
m
−1 −16.0 s
θ = tan
= −79.4◦ ,
3.00 ms
As the vector is in the forth quadrant of the coordinate system the angle
θ = 360◦ − 79.4◦ = 281◦ relative to the positive direction of the x-axis.
4. A plane flies 483 km east from city A to city B in 45.0 min and then 966 km south
from city B to city C in 1.50 h. For the total trip, what are the
(a) magnitude and
(b) direction of the plane’s displacement,
(c) the magnitude and
(d) direction of its average velocity, and
(e) its average speed?
Solutions:
Before we start answering the question let us draw a picture for the situation
described in the question
(a) First let us write down the displacements for the two parts of the trip explicitly:
~r AB = 483 km î
~r BC = −966 km ĵ
Then we get for the displacement ~r AC :
~r AC = ~r AB +~r BC = 483 km î − 966 km ĵ
c 2014 Department of Physics, Eastern Mediterranean University
Page 5 of 10
PHYS101
Motion in two dimensions
Spring 2014
Therefore we get for the magnitude of the displacement, i.e. the trip was
conducted directly from cityA to city C.
q
|~r AC | = r AC = (483 km)2 + (−966 km)2 = 1.08 × 103 km
(b) The direction is then given as:
θ = tan
−1
−966 km
483 km
= −63.4◦
So as you can see also in the figure above the direction of the displacement
is 63.4◦ south of west.
(c) For the average velocity we can use the displacement vector over the time
interval.
~v av =
~r AC
483 km î − 966 km ĵ
483 km î − 966 km ĵ
km
km
=
=
= 215
î − 429
ĵ
∆t
45.0 min + 1.5 h
2.25 h
h
h
So the magnitude of the average velocity is then
s
km
km
km 2
|~v av | = v av =
+ −429
= 480
.
215
h
h
h
(d) The direction of the average velocity is parallel to the displacement vector
and therefore 63.4◦ south of west or 26.6◦ east of south.
(e) Since the average speed is the total distance divided by the total time, we
get for v av.speed
v av.speed =
483 km + 966 km
km
= 644
2.25 h
h
5. A small ball rolls horizontally off the edge of a tabletop that is 1.20 m high. It
strikes the floor at a point 1.52 m horizontally from the table edge.
(a) How long is the ball in the air?
(b) What is its speed at the instant it leaves the table?
Solutions:
First let us draw a picture to understand the situation:
c 2014 Department of Physics, Eastern Mediterranean University
Page 6 of 10
9
10:16
Page 79
PA R T 1
PROBLEMS
PHYS101
he angle between the positive direction
gent to the particle’s path at t ! 2.00 s?
clist is 40.0 m due east of a park’s flaga speed of 10.0 m/s. Then 30.0 s later, the
of the flagpole, going due east with a
yclist in this 30.0 s interval, what are the
ction of the displacement, the (c) magnihe average velocity, and the (e) magnie average acceleration?
ves so that its position (in meters) as
r ! î # 4t2ĵ # tk̂ . Write expresnds) is :
(b) its acceleration as functions of time.
has :
v ! 4.0î " 2.0ĵ # 3.0k̂ and then
" 2.0ĵ # 5.0k̂ (in meters per second). For
roton’s average acceleration :
a avg in unitgnitude of :
a avg, and (c) the angle between
on of the x axis?
e leaves the origin with an initial velocconstant acceleration :
a ! ("1.00î "
es its maximum x coordinate, what are
tion vector?
79
Motion in two dimensions
Spring 2014
point Q on the rim, vertically below P, 0.19 s later. (a) What is the
distance PQ? (b) How far away from the dart board is the dart
released?
y
•22 A small ball rolls horizontally off the edge of a tabletop that
is 1.20 m high. It strikes the floor at a point 1.52 m horizontally
from the table edge. (a) How long is the ball in the air? (b) What is
x
its speed at the instant it leaves the table?
•23 A projectile is fired horizontally from a gun that is
45.0 m above flat ground, emerging from the gun with a speed of
250 m/s. (a) How long does the projectile remain in the air? (b) At
what horizontal distance from the firing point does it strike the
ground? (c) What is the magnitude
of the vertical component of its
h=1.20 m
velocity as it strikes the ground?
•24
In the 1991 World Track and Field Championships in
Tokyo, Mike Powell jumped 8.95 m, breaking by a full 5 cm the 23year long-jump record set by Bob Beamon. Assume that Powell’s
speed on takeoff was 9.5 m/s (about equal to that of a sprinter) and
that g ! 9.80 m/s2 in Tokyo. How much less was Powell’s range
than the maximum possible range for a particle launched at the
same speed?
The current
world-record
is 77.0
set off 1.20 m. As the
(a) We want•25
to calculate
the duration
ofmotorcycle
the balljump
needs
tom,
fall
of a particle moving in the xy plane is
by Jason Renie. Assume that he left the take-off ramp at 12.0º to the
ball is rolling horizontally, there is no vertical component of the initial ve)î # 8.0ĵ , with :
v in meters per second
horizontal and that the take-off and landing heights are the same.
y-direction.
As
illustrated
in the
figure above we put the origin of
) What is the acceleration when tlocity
! 3.0 in Neglecting
air drag,
determine
his take-off
speed.
acceleration zero? (c) When (if ever)
is
the coordinate
system
to the atedge
table,
with
the y-axis
pointing up•26 A stone
is catapulted
time tof
0, with
an initial
velocity
of
! the
n (if ever) does the speed equal 10
m/s? and
magnitude
20.0 m/s
and at antoangle
40.0° above
ward
the x-axis
pointing
the of
right.
Thenthe
wehorizontal.
get for the y-component
ver an xy plane with acceleration compoof the
! "2.0 m/s2. Its initial velocity has com0y ! 12 m/s. In unit-vector notation, what
hen it reaches its greatest y coordinate?
celerates a pebble over a horizontal xy
eration :
a ! (5.00 m/s2)î # (7.00 m/s2)ĵ .
is (4.00 m/s)i.
î What are the (a) magniocity when it has been displaced by 12.0
What arevector:
the magnitudes
position
of the (a) horizontal and (b) vertical
components of its displacement from the catapult site at t ! 1.10
s? Repeat for the (c) horizontal and (d) vertical components at
t ! 1.80 s, and for the (e) horizontal and
1 at
1 (f) vertical components
t ! 5.00
y(t)s. = = y0 + vy t − gt2 = 0 + 0 · t − gt2
⇐⇒
2
2
s airplane has as
••27 ILW A certain
θ
speed of 290.0 km/h and2y
is(diving
t) at
2(−1.20 m)
= 0.495 s
an angle tof %=
below the =
hor- −
! 30.0°−
g
9.80 m
izontal when the pilot releases
a
s2
0
radar decoy (Fig. 4-33). The horia particle moving only on a horizontal
zontalin
distance
between the
The motion
x direction
is arelease
motion under constant velocity. So we can
per seconda is in meters (b)
# 4tĵ , where :
point and the
where
the decoyfrom the horizontal impact position and
conds. At t ! 0, the position easily
vector calculate
thepoint
initial
velocity
strikes the ground is d ! 700 m. (a)
locates the particle, which then has the
the timeHow
the long
ballisspends
theair?
air.
the decoyin
in the
(b)Please recalldthat we set the origin of the
/s)î # (2.00 m/s)ĵ . At t ! 4.00 s, what are
coordinate
edge
of the table, therefore xi = 0
Howsystem
high was at
thethe
release
point?
nit-vector notation and (b) the angle
bel and the
Fig. 4-33 Problem 27.
••28 In Fig. 4-34, a stone is proy
1.52 m − 0
m
x ( t ) − x0
is?
jected
at
a
cliff
of
height
h
x (t) = x0 + v x0 t ⇐⇒ v x0 =
=
= 3.07
with an initial speed of 42.0 m/s directed at
above
A moves
t angle u0 ! 60.0°
0.495
s
s
v
the horizontal. The stone strikes at A, 5.50 s after launching. Find
th a con- A
6. In the figure below,
a stone
a cliff
ofstone
height
h with an initial speed
(a) the height
h of is
theprojected
cliff, (b) the at
speed
of the
just before
itude 3.0
◦
impact
at
A,
and
(c)
the
maximum
height
H
reached
above
the The stone strikes
xis. At the
of 42.0 m/s directed at angle θ0 = 60.0 above the horizontal.
ground.
he y axis,
θ
at A, 5.50 s after launching.
in with a
constant
a
ude 0.40 B
x
a and
en :
Fig. 4-32 Problem 20.
he y axis
A
H
on Analyzed
horizontally with an initial speed of
e bull’s-eye on a dart board. It hits at
θ0
h
Fig. 4-34 Problem 28.
c 2014 Department of Physics, Eastern Mediterranean University
Page 7 of 10
PHYS101
Motion in two dimensions
Spring 2014
Find
(a) the height h of the cliff,
(b) the speed of the stone just before impact at A, and
(c) the maximum height H reached above the ground.
Solutions:
(a) The height of the cliff is y f of the accelerated motion in y-direction. As
depicted in the figure above we put the origin of the coordinate system to
the point where the stone is thrown. The direction of the positive y axis is
pointing upward, and the direction of the positive x-axis is pointing to the
right.
1
h = y(t) = y0 + vy0 t − gt2 =
2
1
m
m
◦
= 0 + 42.0 sin (60 ) 5.50 s − 9.80 2 (5.50 s)2 = 51.8 m
s
2
s
(b) Before we determine the speed, let us determine the velocity in component
notation at the moment of the impact.
m
m
cos (60◦ ) = 21.0
s
s
m
m
m
◦
vy (t) = vy0 − gt = 42.0 sin (60 ) − 9.80 2 5.50 s = −17.5
s
s
s
v x (t) = v x0 = 42.0
Then we get the speed as the magnitude of the velocity vector as
r
q
m 2 m 2
m
2
2
v = v x (t) + vy (t) =
21.0
+ −17.5
= 27.3
s
s
s
(c) At the highest point of the trajectory we have vy = 0 and y = H :
v2y0
42.0 ms sin (60◦ )
H=
=
2g
2 · 9.80 m
s2
2
= 67.5 m
7. A ball is shot from the ground into the air. At a height of 9.1 m, its velocity is
~v = (7.6î + 6.1ĵ)m/s, with î horizontal and ĵ upward.
(a) To what maximum height does the ball rise?
(b) What total horizontal distance does the ball travel?
(c) What are the magnitude and angle (below the horizontal) of the ball’s velocity just before it hits the ground?
c 2014 Department of Physics, Eastern Mediterranean University
Page 8 of 10
PHYS101
Motion in two dimensions
Spring 2014
Solutions:
Given is:
~r0 = 9.1mî,
~v0 = (7.6î + 6.1ĵ)m/s = 0.62s
and
(a) As discussed in the class for constant acceleration (here: ~a = − gĵ), the
motion in x- and y- direction can be considered as independent. So, in order
to determine the maximum height the ball can reach we have to calculate
first the time until it reaches the maximum height, which is defined as the
point where the y-component of the velocity is 0.
vy = vy0 − gT = 0 ⇐⇒ T =
v 0y
g
=
6.1 ms
= 0.62s.
9.8 sm2
Now replacing t into the kinematic equation of the position in y-direction
gives:
1
hmax = y( T ) = y0 + v0y T − gT 2
2
m
1 m
= 9.1m + 6.1 · 0.62s − 9.8 2 (0.62s)2 = 10.99m
s
2 s
So the maximum height h = 11 m.
(b) The position of the ball when it touches the ground is (y0 , R). As y = 0, we
can use the kinematic equations to calculate the range as following:
1
y = y0 + v0y t − gt2 = 0
2
solving this equation for t gives
r
2
− v 0y ±
v 0y − 4 ·
1
2 −2g
t1/2 =
=
v 0y ∓
q
v0y 2 + 2gy0
g
=
− 21 g
y0
=
6.1 ms
∓
q
− v 0y ±
q
v0y 2 + 2gy0
−g
6.1 ms
2
+ 2 · 9.8 sm2 · 9.1m
9.8 sm2
= 2.12s
The mathematical possible negative solution t = −0.88s is not reasonable
and therefore neglected. So, we can now calculate the distance of the ball
from its original position until it hits the ground.
R = x0 + v0x t = 0 + v0x t = 7.6
m
· 2.12s = 16.11m
s
So we get for the range of the ball R = 16.11 m.
c 2014 Department of Physics, Eastern Mediterranean University
Page 9 of 10
PHYS101
Motion in two dimensions
Spring 2014
(c) The velocity in x- direction is constant and therefore v x = 7.6 ms . As we
calculated the time until the impact on the ground as t = 2.12s, we can use
the kinematic equations to determine the y- component of the velocity in
the moment just before it hits the ground.
vy (t) = vy0 − gt = 7.6
m
m
m
− 9.8 2 · 2.12s = −14.7
s
s
s
So, the velocity of the ball just before it hits the ground is then given as:
m
m
~v(t = 2.12s) = 7.6 î − 14.7 ĵ
s
s
So the magnitude of the velocity before it hits the ground is then
r
q
m 2 m 2
m
v = v2x + v2y =
7.6
+ 14.7
= 17
s
s
s
(d) The direction of the ball’s velocity shortly before it hits the ground is then:
θ = tan
−1
−14.7 ms
7.6 ms
= −63◦
So the angle of the ball’s velocity before it hits the ground is 63◦ below the
horizon or 297◦ with respect to the positive x-axis.
c 2014 Department of Physics, Eastern Mediterranean University
Page 10 of 10