In an interval of 10 s, find: a the acceleration, b the final velocity, c the distance moved.. 72 4 Motion in Two Dimensions That is, the displacement vector→r equals the difference betw
Trang 13.7 Exercises 65 (15) Using the formula for the velocity given in the previous exercise, find the
position x of the rocket at any time t during the interval 0 ≤ t ≤ 6 s Then,
find the values of the position, velocity, and acceleration at t = 0, t = 3 s, and
t = 6 s.
(16) A particle has x = 0 at t = 0 and its velocity as a function of time is shown
in the Fig.3.21 (a) Sketch the acceleration as a function of time (b) Find
the average acceleration of the particle in the time interval ti= 0 to tf = 5 s.
(c) Find the acceleration of the particle at t = 4 s.
Fig 3.21 See Exercise (16)
0 2 4 6
-2 -4
t(s)
(17) A particle in one-dimensional motion has a velocity at any instant of time t
given byv = 6 + 4 t + 3 t2 (a) Find the initial velocity when t = 0 (b) Find the
velocity when 2 s have passed (c) Find the expression for the acceleration, and then its value when 2 s have elapsed (d) Find the expression for the displacement
x = x − x◦.
Section 3.5 Constant Acceleration
(18) An object starts from rest and moves with constant acceleration of 4 m/s2 Find
its speed and the distance it has traveled after 5 s have elapsed
(19) A box slides down an incline with a uniform acceleration, see Fig.3.22 It starts from rest and attains a speed of 12 m/s in 4 s Find: (a) the acceleration, and
(b) the distance moved in the first 4 s.
(20) A plane starts from rest and accelerates uniformly along a straight runway before takeoff If the plane moves 1 km in 10 s, then find: (a) the acceleration, (b) the speed at the end of the 10 s period, (c) the distance moved in the first
20 s
Trang 266 3 Motion in One Dimension
Fig 3.22 See Exercise (19)
a
12 m/s
0
t
4 s
t
0
°
a
(21) A particle moving at 25 m/s in a straight line slows uniformly at a rate of 2 m/s every second In an interval of 10 s, find: (a) the acceleration, (b) the final velocity, (c) the distance moved
Section 3.6 Free Fall
(22) A stone strikes the ground with a speed of 25 m/s (a) From what height was it
released? (b) How long was it falling? (c) If the stone is thrown down with a speed of 10 m/s from the same height, then what will be its speed just before hitting the ground?
(23) A ball is thrown upward with a speed of 19.6 m/s (a) How high does it go until
its upward speed decreases to zero? (b) How long does the ball take in this upward trip? (c) How long does the ball take to return to the initial position? (d) What will be its velocity then?
(24) A bottle is dropped from a bridge and strikes the water after 5 s (a) Find the speed of the bottle when it strikes the water (b) Find how high the bridge is located above the water level
(25) A sandbag dropped from a balloon reaches the ground in 5 s, see Fig.3.23 Find the height of the balloon if: (a) it was at rest in the air, (b) it was ascending with
a speed of 10 m/s when the sandbag was dropped, (c) it was descending with a speed of 10 m/s when the sandbag was dropped
(26) A ball is thrown vertically downward from the edge of a cliff with an initial
speed of 23 m/s After a period of 1.4 s has elapsed, find: (a) how fast is it
moving? and (b) how far has it moved?
(27) A ball is thrown vertically upward from the edge of a building with an initial
velocity of 23 m/s After a period of 1.4s has elapsed, find: (a) how fast is it
moving? and (b) how far has it moved?
(28) A ball is thrown vertically upward with a speed of 50 m/s from a building 20 m high, see Fig.3.24 Find:(a) the time t for the ball to reach the highest point,
Trang 33.7 Exercises 67 (b) how high it will rise, (c) how long it will take to return to the starting point, (d) the velocityv2of the ball at this instant, (e) the velocityv3with which the
ball strikes the ground, and (f) the total time of flight t3.
Fig 3.23 See Exercise (25)
Balloon
Fig 3.24 See Exercise (28)
O
(29) A child drops balls from a bridge at regular intervals of 1s, see Fig.3.25 At the moment the fourth ball is released, the first strikes the water (a) How high is the bridge? (b) How far above the water are each of the falling balls at this moment: (c) If the child decided to drop a ball once the previous one has reached the water surface, how long should he wait between every ball drop?
(30) A rubber ball is released from a height of 2 m above the floor, see the Fig.3.26 The ball bounces repeatedly, always rising to 1/2 of the height through which
it falls Treat the ball as a particle that bounces an infinite number of times and
take g= 10 m/s2 (a) What is the average speed of the ball during the first fall?
(b) Show that its total distance traveled during an infinite number of bounces
is 6 m? (c) Show that the total elapsed time for an infinite number of bounces
Trang 468 3 Motion in One Dimension
is 2[√2+ 1]2/√10 s (d) Find the average speed from the time of release to the end of the infinite number of bounces
[Hint: Use the binomial series(1 − x)−1= 1 + x + x2+ x3+ , |x| < 1]
Fig 3.25 See Exercise (29)
Fourth ball Third ball
Second ball
First ball
Fig 3.26 See Exercise (30)
o 1 = 0
o 2
1
(31) An acrobat jumps straight up in the air and his center of mass (CM) took 0.2 s from the moment he just left the ground to the moment he just reached the
Trang 53.7 Exercises 69 highest point, see the Fig.3.27 Neglecting air resistance, (a) what was his initial vertical velocity just before his legs left the ground, (b) how high did his
CM rise above the ground, and (c) what will be his velocity just before touching the ground in his way back?
Fig 3.27 See Exercise (31)
y
y
Before jump
CM
(32) Show that the vertical trajectory of a particle thrown upward is symmetric about its maximum when we neglect air resistance That is, its height above the ground at timet before reaching its maximum equals its height above the
ground after the same time intervalΔt measured after reaching its maximum.
(33) A student drops a dartboard to the ground from a window, i.e.v◦b = 0 One
second after dropping the board, he throws a dart at the board with initial speed
of 20 m/s in order to score just before the board reaches the ground See Fig.3.28
and take g= 10 m/s2 (a) Find the time of flight T of the dartboard (b) Find the
height of the window (c) Find the velocity of both the dart and the dartboard just before hitting the ground
(34) The remote-controlled truck shown in Fig.3.12is used to pick up a package
from a shelf in a factory From rest and at t = 0, the truck accelerate at a1for
a time interval t1, then travels with constant speed for a time interval t2, and
finally decelerate at−a3 for a time interval t3 Show that the total distance traveled by the truck is a1t1(t2+ t3 ) +1
2(a1t12− a3 t32).
(35) A diver drops his body from a diving board at a distance H above the water’s
surface into a deep swimming pool The diver’s motion stops at a distance
h below the surface of the water By choosing the downward direction to be
Trang 670 3 Motion in One Dimension positive, see Fig.3.29, prove that the average acceleration of the diver while he
is under the water is a = −(H/h)g.
o = 20m/s
'b
b
t = T
ob = 0
Fig 3.28 See Exercise (33)
H
mg
mg
F
mg
h
y
w
g
g
o 0
=
v
w=0
v
a
w′
v
Fig 3.29 See Exercise (35)
Trang 7Motion in Two Dimensions 4
This chapter extends the study of the preceding chapter to two dimensions We divide the study into two parts: motion of a particle in a plane, and circular motion of a particle in a plane
The Position Vector
We describe the position of a particle with the position vector→r , which is a vector
that extends from the origin of a certain coordinate system to the particle Using the unit vector notation ofChap 2,→r can be written in two-dimensional form as:
→
where x→
i and y→
j are the vector components of→r along the x and y axes respectively,
and the coefficients x and y are the scalar components, i.e., the particle has the
rectangular coordinates(x, y) In three dimensions the position vector becomes→r =
x→
i + y→j + z→k.
The Displacement Vector
Now, consider a particle moving in the xy plane as shown in Fig.4.1 At point P, let its position be→r
iwhen the time was ti At point Q, let its position be→rf when
the time was tf(the indices i and f refer to the initial and final values for our study) Accordingly, during the time intervalt = tf− ti , the particle’s displacement is:
Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-23026-4_4,
© Springer-Verlag Berlin Heidelberg 2013
Trang 872 4 Motion in Two Dimensions That is, the displacement vector→r equals the difference between the final and initial position vectors As seen from Fig.4.1, the magnitude of the displacement vector is less than the distance traveled along the curved path, which was the particle’s actual path of motion
Fig 4.1 The displacement
→ r = →rf − →ri of a particle
moving in a plane as it moves
from P to Q during the time
intervalt = tf− ti
x
P
Particle’s path
y
Q
i
r
Δr
i
t
f
t
f
r
O
Average Velocity
One of several quantities associated with the phrase “how fast” a particle moves is the average velocity, ¯v, which is defined as follows:
Average velocity
The average velocity,v , of a particle is defined as the ratio of its displacement,→
→r , to the time interval, t That is:
→
v = t→r =→rf−→ri
From this definition,v , has the dimension of length divided by time, that is m/s→
in SI units It is also a vector quantity that has a magnitude and direction along the displacement vector→r
Instantaneous Velocity
Consider the motion of a particle between the two points P and Q in the xy
plane, see Fig.4.2 As the point Q is brought closer and closer to point P (through
points Q1, Q2, ), the time intervals (t1, t2, ) get progressively smaller The
average velocity for each time interval is directed along the displacement vector
As Q approaches P, the time interval approaches zero, and the direction of the
Trang 94.1 Position, Displacement, Velocity, and Acceleration Vectors 73 instantaneous velocity v , which is the direction of the displacement vector,→
approaches the direction of the tangent at P We define v as follows:→
Instantaneous velocity
The instantaneous velocity,→v , of a particle is defined as the limiting value of
the ratio→r /t as t approaches zero, i.e.
→
v = lim
t→0
→r
Fig 4.2 The average velocity
is in the direction of→ r As
Q approaches P, the direction
of→ r and hence the direction
of the instantaneous velocity
→v approaches the tangent line
to the curve at P
x
O
P
Q
y
Direction of at P
Q
1
Q2 Particle’
s path
i
r
i
t
2
r
1
r
f
r
Δr
Δr
1
Δr2
In calculus notation, the above limit is called the derivative of→r with respect to t,
and written as d→r /dt (simplified as ˙→r ) Thus:
→
v = d→r
dt ≡ →rf−→ri=tf
ti
→
From here on, we use the word velocity to designate instantaneous velocity, and speed is defined as the magnitude of that velocity.
In unit vector notation, the position vector can be written in the form→r = x→i + y→j ,
and hence we get:
→
v = d→r
dt = d (x
→
i + y→j ) dt
= d x
dt
→
i +d y
dt
→
j
(4.6)
Trang 1074 4 Motion in Two Dimensions or
→
v = v x
→
where the two components of the velocity vector are given by:
v x = d x
dt ,
v y = d y
dt
(4.8)
Figure4.3shows a velocity vectorv and its scalar components for a particle moving→
in two dimensions
Fig 4.3 The velocity →v
of a particle at point P along
with its scalar components
v xandv y
x
O
P y
r
t
x
article’
s path
We should notice that, in a three dimensional study, the velocity vector can be written in the general form→v = v x→
i + vy→j + vz→k.
Average Acceleration
As the particle moves from P to Q along a certain path in the xy plane as in Fig.4.4, its velocity changes from→viat time tito→vfat time tf.
We define the average acceleration as:
Average acceleration
The average acceleration, a→, of a particle is defined as the ratio of the change
in velocity→v = v→f−→vito the time intervalt = tf− ti That is:
Trang 114.1 Position, Displacement, Velocity, and Acceleration Vectors 75
x
O
P
Q
y
i
r
i
f
r
i
f
f
i
t
f
t
Particle’
s path
Fig 4.4 The average acceleration →a for a particle moving from P to Q is in the direction of the change
in velocity→ v = → vf − →vi shown in the right side of the figure
→
a =→v
t =
→
vf−v→i
Sincet is a scalar quantity, the direction of a→is in the direction of the change in velocity→v =→vf−v→i.
Instantaneous Acceleration
It is useful to define instantaneous acceleration as the limit of the average acceleration whent approaches zero When we consider the motion of the particle between the two points P and Q of the graph shown in Fig.4.4, we see that as point Q approaches
P, the time interval approaches zero, and we define the instantaneous acceleration a→
as follows:
Instantaneous acceleration
The instantaneous acceleration, →a , of a particle is defined as the limiting value
of the ratio v /t when t approaches zero Mathematically→ a→ can be expressed as:
→
t→0
→v
Trang 1276 4 Motion in Two Dimensions
In calculus notation, the above limit is called the first derivative of →v with respect
to t, and written as d v /dt (simplified sometimes as ˙→ →v ), or the second derivative of
→
r with respect to t, and written as d2→r /dt2(simplified sometimes as ¨→r ) Thus:
→
a =d v→
dt = d2→r
dt2 ≡ →vf−→vi=tf
ti
→
From here on, we use the word acceleration to designate instantaneous acceler-ation.
In unit vector notation, we usev = v→ x
→
i + vy→j, so that:
→
a = d→v
dt = d (v x
→
i + vy→j ) dt
= d v x
dt
→
i +d v y
dt
→
j
(4.12)
or
→
a = ax→i + a y→
where the two components of the acceleration vector are given by:
a x = d v x
dt ,
a y = d v y
dt
(4.14)
Figure4.5shows a general view of both the acceleration a→and the velocity →v for
a particle moving in a plane On the same figure, we display the scalar components
of the acceleration vector→a
Fig 4.5 A general view of
the acceleration vector →a of a
particle at point P at a
particular time t The figure
also displays the acceleration
scalar components a x and a y,
as well as the position vector
→r and the velocity vector →v
x
O
P y
r
t
a
x
a
y
a
Pa rticle’
s path
Trang 134.1 Position, Displacement, Velocity, and Acceleration Vectors 77
We should notice that in a three dimensional study, the acceleration vector will take the general form→a = ax→i + ay→j + az→k.
A particle can accelerate for several reasons One way is to change with time the magnitude of the velocity vector (called the speed) such as in one-dimensional motion Another way is to change with time the direction of the velocity vector, as
in circular motion Finally, the acceleration may change due to a change in both the magnitude and the direction of the velocity vector
Example 4.1
A particle moves over a path such that the components of its position with respect
to an origin of coordinates are given as a function of time by:
x = −t2+ 12 t + 5
y = −2 t2+ 16 t + 10
where t is in seconds and x and y are in meters (a) Find the particle’s position
vector →r as a function of time, and find its magnitude and direction at t = 6 s (b)
Find the particle’s velocity vector→v as a function of time, and find its magnitude and direction at t = 6 s (c) Find the particle’s acceleration vector →aas a function
of time, and find its magnitude and direction at t = 6 s.
Solution: (a) The position vector is given at time t by:
→
r = x→i + y→j = (−t2+ 12 t + 5)→i + (−2 t2+ 16 t + 10)→j
Figure4.6shows the variation of x and y as a function of time At t = 6 s we have:
→
r = x→i + y→j = 41→i + 34→j The magnitude of →r is:
r =x2+ y2=412+ 342= 53.26 m
The angleθ between→r and the direction of increasing x is:
θ = tan−1y
x
= tan−1
34 m
41 m
= tan−1(0.83) = 39.7◦
Figure4.7shows the path of the particle in the xy plane, and also shows its position
vector →r at t = 6 s.