Unformatted text preview: CHAPTER 3
Quick Quizzes
1. (c). The largest possible magnitude of the resultant occurs when the two vectors are in the
same direction. In this case, the magnitude of the resultant is the sum of the magnitudes of
A and B: R = A + B = 20 units. The smallest possible magnitude of the resultant occurs
when the two vectors are in opposite directions, and the magnitude is the difference of the
magnitudes of A and B: R = A – B= 4 units. 2. (b). The resultant has magnitude A + B when A is oriented in the same direction as B. 3. (b). The distance traveled, unless you have a very unusual commute, must be larger than
the magnitude of the displacement vector. The distance includes all of the twists and turns
that you made in following the roads from home to work or school. However, the
magnitude of the displacement vector is the length of a straight line from your home to
work or school. The only way that the distance could be the same as the magnitude of the
displacement vector is if your commute is a perfect straight line, which is unlikely! 4. x component y component A – + B + – A+B
5. Vector – – (a) False. As a particle moves in a circle with constant speed, the direction of its velocity
changes continuously, so it undergoes an acceleration.
(b) False. If the velocity is constant, the magnitude of the instantaneous velocity, which is
the speed, is constant. 6. (c). 7. (b). 8. The answers to all three parts are the same—the acceleration is that due to gravity, 9.8
m/s2, because the force of gravity is pulling downward on the ball during the entire
motion. During the rising part of the trajectory, the downward acceleration results in the
decreasing positive values of the vertical component of the velocity of the ball. During the
falling part of the trajectory, the downward acceleration results in the increasing negative
values of the vertical component of the velocity. 53 CHAPTER 3 Problem Solutions
3.1 3.2 Your sketch should be drawn to
scale, and be similar to that
pictured at the right. The angle
from the westward direction,
θ, can be measured as 4°, and the
length of R found to be 7.9 m.
The resultant displacement is
then 7.9 m a t 4° N o f W . N
15.0 meters W
R θ s ter e
0m 3.50
meters 8.2 30.0° S (a) The distance d from A to C is C d = x2 + y2 30 d 0k m φ where x = 200 k m + ( 300 k m ) co s 30.0° = 460 k m B 200 km θ A N and y = 0 + ( 300 km ) sin 30.0° = 150 km . ∴ d = ( 460 km )2 + (150 km )2 = 484 k m
⎛ y⎞
⎛ 150 k m ⎞ 18.1° N of W
(b) φ = ta n −1 ⎜ ⎟ = t a n −1 ⎜
=
⎝ 460 k m ⎟
⎠
⎝ x⎠ 3.3 The displacement vectors A = 8.00 m w est w a r d and
B = 13.0 m n o r t h can be drawn to scale as at the right. The
vector C represents the displacement that the man in the
maze must undergo to return to his starting point. The
scale used to draw the sketch can be used to find C to be
15 m a t 58° S o f E . C 13.0 m W E
8.00 m 3.4 Your vector diagram should look like the
one shown at the right. The initial
displacement A = 100 m d u e w est and the
resultant R = 175 m a t 15.0° N of W are both
known. In order to reach the end point of
the run following the initial displacement,
the jogger must follow the path shown as B.
The length of B and the angle θ can be
measured. The results should be
83 m a t 33° N o f W .
54 S
N R=1
B
W 75 m θ 15.0°
100 m E
S CHAPTER 3.5 3 Using a vector diagram, drawn to scale, like
that shown at the right, the final
displacement of the plane can be found to
= 310 km at θ = 57° N of E . The
be R
plane
requested displacement of the base from
point B is − R plane , which has the same B N m 0k 19 30.0° Rplane magnitude but the opposite direction. Thus,
the answer is A − R plane = 310 k m a t θ = 57° S o f W .
θ 280 km 20.0°
E Base 3.6 (a) Using graphical methods, place the tail of vector B at the head of vector A. The new
vector A + B has a magnitude of 6.1 u n it s a t 113° from the positive xaxis.
(b) The vector difference A − B is found by placing the negative of vector B at the head
of vector A. The resultant vector A − B has magnitude 15 u n it s a t 23° from the
positive xaxis.
y
B A+B –B A A–B x 3.7 (a) In your vector diagram, place the tail of Vector B at the tip
of Vector A. The vector sum, A + B , is then found as
shown in the vector diagram and should be A–B A + B = 5.0 u n it s a t − 53° . –B
+x A (b) To find the vector difference, form the vector –B (same
magnitude, opposite direction) and add it to vector A as
shown in the diagram. You should find that
A − B = 5.0 u n it s a t + 53° . 55 A+B B CHAPTER 3.8 3 (a) Drawing these vectors to scale and maintaining their respective directions yields a
resultant of 5.2 m a t + 60° .
(b) Maintain the direction of A, but reverse the direction of B by 180°. The resultant is
3.0 m a t − 30° .
(c) Maintain the direction of B, but reverse the direction of A. The resultant is
3.0 m a t + 150° .
(d) Maintain the direction of A, reverse the direction of B, and multiply its magnitude
by two. The resultant is 5.2 m a t − 60° .
Using the vector diagram given at the right, we find
R= 6.00 m
θ ( 6.00 m ) + ( 5.40 m ) = 8.07 m
2 2 R ⎛ 5.40 m ⎞
= tan −1 ( 0.900) = 42 .0° .
and θ = ta n −1 ⎜
⎝ 6.00 m ⎟
⎠ E 5.40 m 3.9 S Thus, the required displacement is 8.07 m a t 42.0° S o f E .
3.10 The total displacement is D = 3.10 k m a t 25.0° N o f E . The north and east components of
this displacement are: Dy = ( 3.10 km ) sin 25.0° = 1.31 k m n o r t h ,
and
3.11 Dx = ( 3.10 km ) cos 25.0° = 2.81 k m e a st . (a) Her net x (eastwest) displacement is − 3.00 + 0 + 6.00 = + 3.00 b lo ck s , while her net y
(northsouth) displacement is 0 + 4.00 + 0 = + 4.00 b locks . The magnitude of the
resultant displacement is R= ( x net ) 2 + ( y net ) 2 = ( 3.00) 2 + ( 4.00) 2 = 5.00 blocks , and the angle the resultant makes with the xaxis (eastward direction) is θ = tan −1 ⎛
⎜ 4.00 ⎞
= ta n −1 (1.33) = 53.1°
⎝ 3.00 ⎟
⎠ The resultant displacement is then 5.00 b lock s a t 53.1° N o f E .
56 CHAPTER 3 (b) The total distance traveled is 3.00 + 4.00 + 6.00 = 13.0 b locks .
3.12 + x = ea st w a r d , + y = no r t hw a r d
Σx = 250 m + (125 m ) co s 30.0° = 358 m Σy = 75.0 m + (125 m ) sin 30.0° − 150 m = − 12 .5 m d= ( Σx ) 2 + ( Σy ) = ( 358 m ) 2 + ( −12.5 m ) = 358 m
2 2 ⎛ Σy ⎞
⎛ 12.5 ⎞
= ta n 1 ⎜ –
= − 2 .00° d = 358 m a t 2 .00° S o f E
⎝ 358 ⎟
⎠
⎝ Σx ⎟
⎠ θ = t a n −1 ⎜ A x = −25.0
2
2
A = Ax + Ay = A y = 40.0 y ( −25.0) 2 + ( 40.0) 2 = 47.2 u n it s
40.0 3.13 From the triangle, we find that φ = 58.0° ,
so that θ = 122° .
3.14 A
φ
25.0 θ
x Let A be the vector corresponding to the 10.0 yd run, B to the 15.0 yd run, and C to the
50.0 yd pass. Also, we choose a coordinate system with the +y direction downfield, and
the + x direction toward the sideline to which the player runs.
The components of the vectors are then
A y = − 10.0 yd s
Ax = 0
Bx = 15.0 y d s By = 0 Cx = 0 Cy = + 50.0 yd s From these, R x = ΣFx = 15.0 y d s , and Ry = ΣFy = 40.0 yd s ,
and 2
2
R = Rx + Ry = (15.0 y d s ) + ( 40.0 yd s )
2 57 2 = 42.7 y a r d s . CHAPTER 3.15 3 After 3.00 h moving at 41.0 km/h, the hurricane is 123 km at 60.0° N of W from the
island. In the next 1.50 h, it travels 37.5 km due north. The components of these two
displacements are:
Displacement
123 km
37.5 km
Resultant xcomponent (eastward)
61.5 km
0
61.5 km ycomponent (northward)
+107 km
+37.5 km
144 km Therefore, the eye of the hurricane is now R= 3.16 2
(− 61.5 km ) + (144 km ) 2 = 157 k m fr om t h e isla n d . Choose the positive x direction to be eastward and positive y as northward. Then, the
components of the resultant displacement from Dallas to Chicago are:
R x = Σx = ( 730 m i ) co s 5.00° − ( 560 m i ) sin 21.0° = 527 m i , and Ry = Σy = ( 730 m i ) sin 5.00° + ( 560 m i ) cos 21.0° = 586 m i .
2
2
R = Rx + R y = ( 527 m i ) 2 + ( 586 m i ) 2 = 788 m i Σy ⎞
= t a n −1 (1.11) = 48.1°
⎝ Σx ⎟
⎠ θ = t a n −1 ⎜ Thus, the displacement from Dallas to Chicago is R = 788 m i a t 48.1° N of E . 58 CHAPTER 3.17 3 The components of the displacements a, b, and c are:
ax = a ⋅ co s 30.0° = + 152 k m , y north
c bx = b ⋅ cos 110° = − 51.3 km ,
cx = c ⋅ co s 180° = − 190 k m ,
and b
R ay = a ⋅ sin 30.0° = + 87.5 km , θ 20.0°
110° a
30.0° by = b ⋅ sin 110° = + 141 km , x east cy = c ⋅ sin 180° = 0 .
Thus, Rx = ax + bx + cx = − 89.7 km , and Ry = ay + by + cy = + 228 km ,
⎛R ⎞
2
2
R = R x + Ry = 245 km , and θ = tan −1 ⎜ x ⎟ = t a n −1 (1.11) = 21.4°
⎝ Ry ⎠
City C is 245 k m a t 21.4° W o f N from the starting point. so 3.18 F = 120 N
1 F x = 60.0 N
1 F y = 104 N
1 F2 = 80.0 N (a) F2 x = − 20.7 N F2 y = 77.3 N R= ( ΣFx ) 2 + ( ΣFy ) 2 = ( 39.3 N ) 2 + (181 N ) 2 = 185 N ⎛ 181 N ⎞
= tan −1 ( 4.61) = 77.8°
and θ = t an −1 ⎜
⎝ 39.3 N ⎟
⎠
The resultant is R = 185 N a t 77.8° fr o m t h e xa xis .
(b) To have zero net force on the mule, the resultant above must be cancelled by a force
equal in magnitude and oppositely directed. Thus, the required force is
185 N a t 258° fr o m t h e xa xis . 59 CHAPTER 3.19 3 The resultant displacement is R = A + B , where A is the 150 cm displacement at 120° and
B is the required second displacement. Solving for B: B = R − A = R + ( − A ) .
The components of B are
Bx = R x − A x = + 190 cm and By = Ry − A y = − 49.6 cm .
⎛ By ⎞
2
2
Hence, B = Bx + By = 196 cm and θ = tan −1 ⎜ ⎟ = tan −1 ( − 0.262) = − 14.7° .
⎝ Bx ⎠
B = 196 cm a t 14.7° b elo w t h e xa xis . 3.20 Let +x = East, +y = North,
Displacement
300 km, due E
350 km, 30° W of N
150 km, due N
Resultant x (km)
300
–175
0
Σx = 125 ⎛ Σy ⎞
(a) θ = tan −1 ⎜ ⎟ = 74.6° N of E
⎝ Σx ⎠ y (km)
0
303
150
Σy = 453
(b) 60 R= 2
( Σx )2 + ( Σy ) = 470 km CHAPTER 3.21 3 (a) Your first displacement takes you to point A, so r1 = rA . In the second displacement,
1
you go onehalf the distance from A toward B, so ∆r= ( rB − rA ) and your current
2
1
rA + rB
. On the next leg of the hunt,
position vector is r2 = r1 + ∆r= rA + ( rB − rA ) =
2
2
1
your displacement is ∆r=rA + ( rC − r2 ) and your new position vector becomes
3
r +r +r
1
2
1
r3 = r2 + ∆r = r2 + ( rC − r2 ) = r2 + rC = A B C . The next displacement is
3
3
3
3
1
∆r= ( rD − r3 ) and your position vector changes to
4
r +r +r +r
1
3
1
r4 = r3 + ∆r = r3 + ( rD − r3 ) = r3 + rD = A B C D . On the final leg of the hunt,
4
4
4
4
1
the displacement is ∆r= ( rE − r4 ) . Therefore, the position vector of the treasure is
5
1
r5 = r4 + ∆r = r4 + ( rE − r4 ) .
5
r +r +r +r +r
4
1
= r4 + rE = A B C D E .
5
5
5 To determine the coordinates of this location, we consider
R = rA + rB + rC + rD + rE . Vector
rA
rB
rC
rD
rE R x component (m)
30.0
60.0
–10.0
40.0
–70.0 Σy = 80.0 Σx = 50.0 The position vector of the treasure is r5 =
x= y component (m)
–20.0
80.0
–10.0
–30.0
60.0 R
and its coordinates are then seen to be
5 1
1
R x = + 10.0 m and y = R y = + 16.0 m .
5
5 (b) From the solution of part (a), the position vector of the treasure’s location is seen to
depend on the sum of the position vectors of the individual trees:
R r +r +r +r +r
r5 = = A B C D E . Interchanging the trees would only change the order of
5
5
the vectors in this sum. Since a vector sum is independent of the order in which the
vectors are added, the answer found in part (a) does not depend on the order of the
trees.
61 CHAPTER 3.22 3 v ix = 100.8 m i h = 45.06 m s and ∆x = 60.0 ft = 18.3 m The time to reach home plate is t = ∆x
18.3 m
=
= 0.406 s .
v ix 45.06 m s In this time interval, the vertical displacement is ( ) 1
1
2
∆y = v iy t + ay t 2 = 0 + −9.80 m s 2 ( 0.406 s ) = − 0.807 m .
2
2 ⎛ 3.281 ft ⎞
= 2.65 ft .
Thus, the ball drops vertically 0.807 m ⎜
⎝ 1m ⎟
⎠
3.23 The constant horizontal speed of the falcon is v x = 200 mi ⎛ 0.447 m
h ⎜ 1 mi h
⎝ s⎞
⎟ = 89.4 m s .
⎠ The time required to travel 100 m horizontally is t = ∆x
100 m
=
= 1.12 s . The vertical
v x 89.4 m s displacement during this time is ( ) 1
1
2
∆y = v iy t + ay t 2 = 0 + −9.80 m s 2 (1.12 s ) = −6.13 m ,
2
2 or the falcon has a vertical fall of 6.13 m . 3.24 1
We find the time of fall from ∆y = v iy t + ay t 2 with v iy = 0 :
2
t= 2 ( ∆y )
a = 2 ( − 50.0 m )
− 9.80 m s 2 = 3.19 s . At impact, v x = v ix = 18.0 m s , and the vertical component is ( ) v y = v iy + ay t = 0 + − 9.80 m s 2 ( 3.19 s ) = − 31.3 m s . 62 CHAPTER (18.0 2
2
Thus, v = v x + v y = 3 m s ) + ( −31.3 m s ) = 36.1 m s
2 2 ⎛ vy ⎞
−1 ⎛ −31.3 ⎞
⎟ = t a n ⎜ 18.0 ⎟ = −60.1° ,
⎝
⎠
⎝ vx ⎠ and
or
3.25 θ = ta n −1 ⎜ v = 36.1 m s a t 60.1° b elo w t h e h o r iz o n t a l . At the maximum height v y = 0 , and the time to reach this height is found from v y = v iy + ay t as t = v y − v iy
ay = 0 − v iy
−g = v iy
g . The vertical displacement that has occurred during this time is
2
⎛ v y + v iy ⎞ ⎛ 0 + v iy ⎞ ⎛ v iy ⎞ v iy
∆y ) max = v y t = ⎜
=⎜
=
.
t
(
⎝ 2 ⎟ ⎝ 2 ⎟ ⎜ g ⎟ 2g
⎠
⎠⎝ ⎠ ⎛ 1m ⎞
= 3.7 m , then
Thus, if ( ∆y ) max = 12 ft ⎜
⎝ 3.281 ft ⎟
⎠ ( ) v iy = 2 g ( ∆y ) max = 2 9.80 m s 2 ( 3.7 m ) = 8.5 m s ,
and if the angle of projection is θ = 45° , the launch speed is vi = 3.26 v iy
sin θ = 8.5 m s
= 12 m s .
sin 45° 1
The time of flight for Tom is found from ∆y = v iy t + ay t 2 with v iy = 0 :
2
t= 2 ( ∆y )
a = 2 ( − 1.5 m )
− 9.80 m s 2 = 0.55 s . The horizontal displacement during this time is
∆x = v ix t = ( 5.0 m s ) ( 0.55 s ) = 2 .8 m . Thus, he lands 2 .8 m fr om t h e ba se o f t h e ta ble . 63 CHAPTER 3 The horizontal component of velocity does not change during the flight, so
v x = v ix = 5.0 m s . The vertical component of velocity is found as ( ) v y = v iy + ay t = 0 − − 9.80 m s 2 ( 0.55 s ) = − 5.4 m s . 3.27 When ∆y = ( ∆y )max , v y = 0 .
Thus, v y = v iy + ay t yields 0 = v i sin 3.00° − gt or t = v i sin 3.00°
.
g 1
The vertical displacement is ∆y = v iy t + ay t 2 . At the maximum height, this becomes
2 ( ∆y )max 2 ⎛ v sin 3.00° ⎞ 1 ⎛ v i sin 3.00° ⎞
v 2 sin 2 3.00°
.
= ( v i sin 3.00°) ⎜ i
− g⎜
=i
⎟ 2⎝
⎟
g
g
2g
⎝
⎠
⎠ If ( ∆y )max = 0.330 m , the initial speed is vi = 2 g ( ∆y ) max
sin 3.00° = ( ) 2 9.80 m s 2 ( 0.330 m )
sin 3.00° = 48.6 m s . Note that it was unnecessary to use the horizontal distance of 12.6 m in this solution.
3.28 The horizontal displacement at t = 42 .0 s is
3
x = v ix t = ( v i co s θ ) t = ( 300 m s ) ( co s 55.0°) ( 42.0 s ) = 7.23 × 10 m . The vertical displacement at t = 42 .0 s is
1
1
y = v iy t + ay t 2 = ( v i sin θ ) t − gt 2
2
2
= ( 300 m s ) ( sin 55.0°) ( 42.0 s ) + ( ) 1
2
3
−9.80 m s 2 ( 42.0 s ) = 1.68 × 10 m
2 64 CHAPTER 3.29 3 We choose our origin at the initial position of the projectile. After 3.00 s, it is at ground
level, so the vertical displacement is ∆y = − H .
1
To find H, we use ∆y = v iy t + ay t 2 , which becomes
2
− H = (15 m s ) ( sin 25°) ( 3.0 s ) + 3.30 ( ) 1
2
−9.80 m s 2 ( 3.0 s ) , or H = 25 m .
2 The components of the initial velocity are:
v ix = v i co s 53.0° = 12 .0 m s , and v iy = v i sin 53.0° = 16.0 m s . (a) The time required for the ball to reach the position of the crossbar is
t= ∆x
36.0 m
=
= 3.00 s .
v ix 12 .0 m s At this time, the height of the football above the ground is 1
m⎞
1⎛
m⎞
2
⎛
∆y = v iy t + ay t 2 = ⎜ 16.0 ⎟ ( 3.00 s ) + ⎜ − 9.80 2 ⎟ ( 3.00 s ) = 3.90 m
⎝
⎠
⎝
⎠
2
s
2
s
Thus, the ball clea r s t h e cr o ssb a r b y 3.90 m − 3.05 m = 0.85m .
(b) The vertical component of the velocity of the ball as it moves over the crossbar is
v y = v iy + ay t = (16.0 m s ) + − 9.80 m s 2 ( 3.00 s ) = − 13.4 m s . The negative sign ( ) indicates the ball is moving downward or fa llin g .
3.31 The speed of the car when it reaches the edge of the cliff is ( ) v = v i2 + 2a( ∆x ) = 0 + 2 4.00 m s 2 ( 50.0 m ) = 20.0 m s .
Now, consider the projectile phase of the car’s motion. The vertical velocity of the car as
it reaches the water is ( ) v y = − v i2y + 2ay ( ∆y ) = ⎡ − ( 20.0 m s ) sin 24.0°⎤ + 2 − 9.80 m s 2 ( − 30.0 m ) ,
⎣
⎦
2 or v y = − 25.6 m s . 65 CHAPTER 3 (b) The time of flight is
t= v y − v iy
ay = − 25.6 m s − ⎡ − ( 20.0 m s ) sin 24.0°⎤
⎣
⎦ = 1.78 s .
− 9.80 m s 2 (a) The horizontal displacement of the car during this time is ∆x = v ix t = ⎡( 20.0 m s ) cos 24.0°⎤ (1.78 s ) = 32 .5 m .
⎣
⎦
3.32 The components of the initial velocity are
v ix = ( 40.0 m s ) co s 30.0° = 34.6 m s , and
v iy = ( 40.0 m s ) sin 30.0° = 20.0 m s . /s
0 m 30.0°
0. y 4 The time for the water to reach the building is
t= 50.0 m ∆x 50.0 m
=
= 1.44 s .
v ix 34.6 m The height of the water at this time is ( ) 1
1
2
∆y = v iy t + ay t 2 = ( 20.0 m s ) (1.44 s ) + − 9.80 m s 2 (1.44 s ) = 18.6 m .
2
2 3.33 (a) At the highest point of the trajectory, the projectile is moving horizontally with
velocity components of v y = 0 and
v x =v ix =v i co s θ = ( 60.0 m / s ) cos 30.0° = 52.0 m s . (b) The horizontal displacement is ∆x = v ix t = ( 52.0 m s ) ( 4.00 s ) = 208 m and, from
1
∆y = ( v i sin θ ) t + ay t 2 , the vertical displacement is
2
∆y = ( 60.0 m s ) ( sin 30.0°) ( 4.00 s ) + ( ) 1
2
−9.80 m s 2 ( 4.00 s ) = 41.6 m .
2 The straight line distance is d= ( ∆x ) 2 + ( ∆y ) = ( 208 m ) 2 + ( 41.6 m ) 2 =
2 66 212 m . CHAPTER 3.34 3 The horizontal kick gives zero initial vertical velocity to the ball. Then, from
1
∆y = v iy t + ay t 2 , the time of flight is
2
t= 2 ( ∆y )
a = 2 ( − 40.0 m )
− 9.80 m s 2 = 8.16 s . The extra time ∆t = 3.00 s − 8.16 s = 0.143 s is the time required for the sound to travel
in a straight line back to the player. The distance the sound travels is d= ( ∆x ) 2 + ( ∆y ) = v sound ∆t
2 where ∆x represents the horizontal displacement of the ball when it hits the water. Thus,
∆x = d 2 − ( ∆y ) = ⎡( 343 m s ) ( 0.143 s ) ⎤ − ( − 40.0 m ) = 28.3 m .
⎣
⎦
2 2 2 The initial velocity given the ball must have been ∆x 28.3 m
=
= 9.91 m s
t
8.16 s v i = v ix = 3.35 The velocity of the plane relative
north
to the ground is the vector sum of
the velocity of the plane relative
to the air and the velocity of the
air relative to the ground, or v pg = v pa + v ag . vpg
θ
vpa = 300 mph The components of this velocity are v pg east = 300 m i h + (100 m i h ) cos30.0° = 387 m i h ,
= 0 + (100 m i h ) sin 30.0° = 50.0 m i h . and v Thus, v 2
2
= ⎛v ⎞
+ ⎛v ⎞
=
pg
⎝ pg ⎠ east ⎝ pg ⎠ n ort h and θ = tan −1 ⎜ pg north ⎛ v pg
⎜ v pg
⎝ nort h
east ( 387 )2 + ( 50.0) 2 m i h = 390 m i h ⎞
⎛ 50.0 ⎞
⎟ = tan −1 ⎜
= 7.37°
⎝ 387 ⎟
⎠
⎟
⎠ The plane moves at 390 m i h a t 7.37° N o f E r ela t iv e t o t h e g r o u n d .
67 vag = 100 mph
30.0°
east CHAPTER 3.36 3 We use the following notation:
v bs = velocity of boat relative to the shore
v bw = velocity of boat relative to the water,
and
v ws = velocity of water relative to the shore.
If we take downstream as the positive direction, then v ws = + 1.5 m s for both parts of
the trip. Also, v bw = + 10 m s while going downstream and v bw = − 10 m s for the
upstream part of the trip.
The velocity of the boat relative to the shore is given by v bs = v bw + v w s .
While going downstream, v bs = 10 m s + 1.5 m s and the time to go 300 m downstream
is t dow n = 300 m
= 26 s .
(10+1.5) m s When going upstream, v bs = − 10 m s + 1.5 m s = − 8.5 m s and the time required to move
− 300 m
300 m upstream is t up =
= 35 s .
− 8.5 m s
The time for the round trip is t = t down + tup = ( 26 + 35) s = 61 s .
3.37 The velocity of the plane relative to the ground
is the vector sum of the velocity of the plane
relative to the air and the velocity of the air
relative to the ground as shown in the diagram.
Thus, v pg =
and north
vpg
30.0 km/h
west θ
150 km/h (150) 2 + ( 30.0) 2 km h = 153 km h
θ = tan −1 ⎛
⎜ 30.0 ⎞
= 11.3° .
⎝ 150 ⎟
⎠ The plane has a velocity of 153 k m h a t 11.3° N o f W relative to the ground. 68 CHAPTER 3.38 3 v bw = 10 m s , directed northward, is the velocity of the boat relative
to the water. north v ws = 1.5 m s , directed eastward, is the velocity of the water relative
to shore. vbw θ
vbs
east v bs is the velocity of the boat relative to shore, and directed at an angle
of θ, relative to the northward direction as shown. vws v bs = v bw + v w s The northward component of v bs is v bs co s θ = v bw = 10 m s . (1) v bs sin θ = v ws = 1.5 m s . (2) The eastward component is (a) Dividing equation (2) by equation (1) gives
⎛ v ws ⎞
−1 ⎛ 1.50 ⎞
⎟ = t a n ⎜ 10.0 ⎟ = 8.53° .
⎝
⎠
⎝ v bw ⎠ θ = t a n −1 ⎜ From equation (1), v bs = 10 m s
= 10.1 m s .
co s 8.53° Therefore, v bs = 10.1 m s a t 8.53° E o f N .
300 m
300 m
=
= 30.0 s and the downstream
v bs co s θ 10.0 m s
drift of the boat during this crossing is (b) The time to cross the river is t = drift = ( v bs sin θ ) t = (1.50 m s ) ( 30.0 s ) = 45.0 m . 3.39 v bw = velocity of boat relative to the water,
v ws = velocity of water relative to the shore and v bs = velocity of boat relative to the shore.
v bs = v bw + v w s as shown in the diagram. vbs
vbw
62.5° The northward (i.e., crossstream) component of vbs is ( v bs )north = ( v bw ) sin 62 .5° + 0 = ( 3.30 m i h ) sin 62 .5° + 0 = 2 .93 m i h . 69 north vws east CHAPTER The time required to cross the stream is then t = 3 0.505 m i
= 0.173 h .
2.93 m i h The eastward (i.e., downstream) component of v bs is ( v bs ) east = − ( v bw ) co s 62 .5° + v ws
= − ( 3.30 m i h ) co s 62 .5° + 1.25 m i h = − 0.274 m i h . Since the last result is negative, it is seen that the boat moves upstream as it crosses the
river. The distance it moves upstream is ⎛ 5280 ft ⎞
d = ( v bs ) east t = ( 0.274 m i h ) ( 0.173 h ) = 4.72 × 10−2 m i ⎜
= 249 ft .
⎝ 1 mi ⎟
⎠ ( 3.40 ) v pa = the velocity of the plane relative to the air N = 200 k m h . v ag = the velocity of the air relative to the ground
= 50.0 k m h (south). W vpa
θ
vpg v pg = the velocity of the plane relative to the ground
(to be due west).
(a) The velocity of the plane relative to the ground is given by the vector sum
v pg = v pa + v ag . If v pg is to have zero northward component as shown in the
diagram, we must have v pa sin θ = v ag , or
⎛ v ag ⎞
−1 ⎛ 50.0 km h ⎞
= 14.5° .
⎟ = s in ⎜
⎝ 200 k m h ⎟
⎠
⎝ v pa ⎠ θ = sin −1 ⎜ Thus, the plane should head at 14.5° N o f W .
(b) Since v pg has zero northward component, the plane’s ground speed is
v pg = v pa co s θ = ( 200 k m h ) co s 14.5° = 194 km h . 70 E
vag
S CHAPTER 3.41 3 The velocity of the faster car relative to the slower car is given by v fs = v fe + v es , where v fe = + 60.0 km h is the velocity of the faster car relative to Earth and
v es = − v se = − 40.0 k m h is the velocity of Earth relative to the slower car. Thus, v fs = + 60.0 km h − 40.0 km h = + 20.0 km h and the time required for the faster
car to move 100 m (0.100 km) closer to the slower car is t= 3.42 d
0.100 km
⎛ 3600 s ⎞ 18.0 s
=
= 5.00 × 10−3 h ⎜
⎟=
v fs 20.0 km h
⎝ 1h ⎠ v bc = the velocity of the ball relative to the car.
v ce = velocity of the car relative to the earth = 10 m s .
v be = the velocity of the ball relative to the earth. These velocities are related by the equation v be = v bc + v ce
as illustrated in the diagram. vbc
vbe
60.0°
vce Considering the horizontal components, we see that
v bc co s 60.0° = v ce or v bc = 10.0 m s
v ce
=
= 20.0 m s .
co s 60.0° co s 60.0° From the vertical components, the initial velocity of the ball relative to the earth is
v be = v bc sin 60.0° = 17.3 m s .
2
Using v y = v i2y + 2 ay ( ∆y ) , with v y = 0 when the ball is at maximum height, we find ( ∆y )max = 2
0 − v iy 2 ay 2
(17.3 m s ) = 15.3 m
0 − v be
=
=
2 ( − g ) 2 9.80 m s 2
2 ( ) as the maximum height the ball rises. 71 CHAPTER 3.43 3 Since R = A + B , then B = R − A and the components of the
second displacement are: y
θ
B Bx = R x − A x A R 120° 35.0° = (140 cm ) cos 35.0° − (150 cm ) cos 120° = + 190 cm x and By = Ry − A y = (140 cm ) sin 35.0° − (150 cm ) sin 120° = − 49.6 cm .
⎛ By ⎞
2
2
Thus, B = Bx + By = 196 cm , and θ = t a n −1 ⎜ ⎟ = t a n −1 ( −0.262) = −14.7° .
⎝ Bx ⎠ The second displacement is 196 cm a t 14.7° b e lo w t h e p o sit iv e x a xis .
3.44 Observe that when one chooses the x and y axes
as shown in the drawing, each of the four forces
lie along one of the axes. The resultant, R, is easily
computed as y
R
x 31.0 N θ 12.0 N
35.0° Rx = ΣFx = +12 .0 N − 8.40 N = +3.60 N
8.40 N horizontal Ry = ΣFy = +31.0 N − 24.0 N = +7.00 N
24.0 N
2
2
R = R x + Ry = ( 3.60 N ) 2 + ( 7.00 N ) 2 = 7.87 N ⎛ Ry ⎞
= 62.8° , or 62.8° + 35.0°= 97.8° fr o m t h e h o r iz o n t a l .
⎝ Rx ⎟
⎠ θ = t a n −1 ⎜ R = 7.87 N a t 97.8° co u n t er clo ck w ise fr o m t h e h o r iz o n t a l lin e t o t h e r ig h t . 3.45 v ce = the velocity of the car relative to the earth.
v wc = the velocity of the water relative to the car.
v we = the velocity of the water relative to the earth. These velocities are related as shown in the diagram at the right. 72 vce
vwe 60.0° vwe = vce + vwc vwc CHAPTER 3 (a) Since vwe is vertical, v wc sin 60.0° = v ce = 50.0 k m h
or v wc = 57.7 k m h a t 60.0° w est o f v er t ica l . (b) Since v ce has zero vertical component,
v we = v w c co s 60.0° = ( 57.7 k m h ) co s 60.0° = 28.9 k m h d o w n w a r d . 3.46 (a) v ix = v i co s θ = (10.0 m s ) co s 37.0° = 7.99 m s
v iy = v i sin θ = (10.0 m s ) sin 37.0° = 6.02 m s /s
m
.0
10
i=
37.0°
v (b) First, we make sure the stone does not hit the
front edge of the dock.
When ∆x = 3.00 m , t = 3.00 m ∆x
3.00 m
=
= 0.376 s .
v ix 7.99 m s At this time, ( ) 1
2
∆y = v iy t + ay t 2 = ( 6.02 m s ) ( 0.376 s ) − 4.90 m s 2 ( 0.376 s )
2 or ∆y = 1.57 m > 1.00 m . The stone clears the front corner of the dock.
2
To find the maximum height reached, use v y = v i2y + 2ay ( ∆y ) . Since v y = 0 w h en ∆y = ( ∆y ) max , this gives ( ∆y )max = 2
0 − v iy 2 ay = − ( 6.02 m s ) ( 2 2 − 9.80 m s 2 ) = 1.85 m a bov e th e gr ou n d . 73 1.00 m CHAPTER 3 1
(c) When ∆y = 1.00 m , ∆y = v iy t + ay t 2 yields
2
1.00 m = ( 6.02 m s ) t + ( ( ) 1
−9.80 m s 2 t 2
2 ) or t 2 − (1.23 s ) t + 0.204 s 2 = 0 with solutions of
t = 0.198 s and t = 1.03 s.
Since the stone hits the dock as it comes back down, use the second solution
t = 1.03 s. At this time, ∆x = v ix t = ( 7.99 m s ) (1.03 s ) = 8.23 m .
Thus, the stone hits the dock 5.23 m b eyon d t h e e d ge .
(d) At t = 1.03 s, ( ) v y = v iy + ay t = 6.02 m s − 9.80 m s 2 (1.03 s ) = − 4.08 m s
v x = v ix = 7.99 m s . and 2
2
Therefore, v = v x + v y = 8.97 m s 3.47 AC = v 1t = ( 90.0 km h ) ( 2.50 h ) = 225 k m C BD = AD − AB = A C co s 40.0° − 80.0 k m = 92.4 k m From the triangle BCD, BC =
= ()()
2 BD + DC 40.0°
A 2 ( 92.4 km ) 2 + ( AC sin 40.0°) =172 km
2 Since Car 2 travels this distance in 2.50 h, its constant speed is
v2 = 172 k m
= 68.6 k m h
2.50 h 74 80.0 km B D CHAPTER 3.48 3 The three diagrams shown below represent the graphical solutions for the three vector
sums: R 1 = A + B + C , R 2 = B + C + A , a n d R 3 = C + B + A . You should observe that
R 1 = R 2 = R 3 , illustrating that the sum of a set of vectors is not affected by the order in
which the vectors are added.
C B A A B
R1 R2 A C R3 C B 3.49 The distance, s, moved in the first 3.00 seconds is given by ( ) 1
1
2
s = v i t + at 2 = (100 m s ) ( 3.00 s ) + 30.0 m s 2 ( 3.00 s ) = 435 m .
2
2 At the end of powered flight, the coordinates of the rocket are:
x 1 = s co s 53.0° = 262 m , a n d y 1 = s sin 53.0° = 347 m The speed of the rocket at the end of powered flight is ( ) v 1 = v i + at = 100 m s + 30.0 m s 2 ( 3.00 s ) = 190 m s ,
so the initial velocity components for the freefall phase of the flight are
v ix = v 1 co s 53.0° = 114 m s and v iy = v1 sin 53.0° = 152 m s .
(a) When the rocket is at maximum altitude, v y = 0 . The rise time during the freefall
phase can be found from v y = v iy + ay t as
t rise = 0 − v iy
ay = 0 − 152 m
= 15.5 s .
− 9.80 m s 2 The vertical displacement occurring during this time is ⎛ v fy + v iy ⎞
⎛ 0 + 152 m s ⎞
3
∆y = ⎜
⎟ (15.5 s ) = 1.17 × 10 m .
⎟ t rise = ⎜
⎝
⎠
2⎠
2
⎝
The maximum altitude reached is then H = y1 + ∆y = 347 m + 1.17 × 103 m = 1.52 × 103 m
75 CHAPTER 3 (b) After reaching the top of the arc, the rocket falls 1.52 × 103 m to the ground, starting ( ) with zero vertical velocity v iy = 0 . The time for this fall is found from
1
∆y = v iy t + ay t 2 as
2
t fall = 2 ( ∆y )
ay = ( 2 −1.52 × 10 3 m
9.80 m s 2 ) = 17.6 s . The total time of flight is t = t powered + trise + t fall = ( 3.00 + 15.5 + 17.6) s = 36.1 s .
(c) The freefall phase of the flight lasts for t 2 = trise + t fall = (15.5 + 17.6) s = 33.1 s .
The horizontal displacement occurring during this time is
∆x = v ix t 2 = (114 m s ) ( 33.1 s ) = 3.78 × 10 3 m and the full horizontal range is
3
R = x1 + ∆x = 262 m + 3.78 × 103 m = 4.05 × 10 m . 3.50 The velocity of a canoe relative to the shore is given by v cs = v cw + v w s , where v cw is the
velocity of the canoe relative to the water and v ws is the velocity of the water relative to
shore.
Applied to the canoe moving upstream, this gives
− 1.2 m s = − v cw + v ws (1) and for the canoe going downstream
+2 .9 m s = + v cw + v ws (2) (a) Adding equations (1) and (2) gives
2 v ws = 1.7 m s , so v w s = 0.85 m s . 76 CHAPTER 3 (b) Subtracting (1) from (2) yields
2 v cw = 4.1 m s , or v cw = 2.1 m s . 3.51 2v iy
1
The time of flight is found from ∆y = v iy t + ay t 2 with ∆y = 0 , as t =
. This gives the
2
g
2v ix v iy
.
range as R = v ix t =
g
On Earth this becomes REarth = 2v ix v iy
gEart h , and on the moon, R M oon = 2v ix v iy
g M oon . ⎛g
⎞
⎛ 1⎞
Dividing R M oon by R Eart h , we find R M oon = ⎜ Eart h ⎟ REarth . With g M oon = ⎜ ⎟ gEarth , this gives
⎝ 6⎠
⎝ g M oon ⎠
R M oon = 6REart h = 6 ( 3.0 m ) = 18 m .
⎛g
⎞
3.0 m
= 7.9 m .
Similarly, R M ars = ⎜ Eart h ⎟ REart h =
0.38
⎝ g Mars ⎠ 3.52 The time to reach the opposite side is t = ∆x
10 m
.
=
v ix v i co s 15° When the motorcycle returns to the original level, the vertical displacement is ∆y = 0 .
1
Using this in the relation ∆y = v iy t + ay t 2 gives a second relation between the takeoff
2
speed and the time of flight as:
0 = ( v i sin 15°) t + ⎛
g⎞
1
( − g ) t 2 or v i = ⎜ 2 sin 15° ⎟ t .
2
⎝
⎠ Substituting the time found earlier into this result yields the required takeoff speed as ( 9.80 m s ) (10 m ) =
2 vi = 2 ( sin 15°) ( cos 15°) 14 m s . 77 CHAPTER 3.53 1
The time to fall 36.0 m, starting with v iy = 0 , is found from ∆y = v iy t + ay t 2 as
2
1
−36.0 m = 0 + −9.80 m s 2 t 2 , yielding t = 2 .71 s . Hence, the minimum initial horizontal
2
∆x 6.00 m
speed is v ix min =
=
= 2.21 m s .
t
2.71 s ( 3.54 3 ) For a projectile fired with initial speed v i at angle θ above the horizontal, the time
∆x
∆x
required to achieve a horizontal displacement of ∆x is t =
. The vertical
=
v ix v i co s θ
displacement of the projectile at this time is ⎛ ∆x ⎞ g ⎛ ∆x ⎞
g ( ∆x )
1
∆y = v iy t + ay t 2 = ( v i sin θ ) ⎜
⎟ − 2 ⎜ v cos θ ⎟ = ( ∆x ) t a n θ − 2 cos 2 θ v 2
2
⎝ v i cos θ ⎠
⎝i
⎠
i
2 2 ( Solving for the initial speed gives v i = g ( ∆x ) 2 ( 2cos θ ) ⎡( ∆x ) tan θ − ∆y ⎤
⎣
⎦
2 . With ∆x = 50 m , ∆y = 30 m , θ=55°, an d g = 9.80 m s 2 ,
we find the required initial speed to be v i = 30 m s . 3.55 (a) The time to reach the fence is t = ∆x
130 m
159 m
=
=
.
v ix v i cos 35°
vi At this time, the ball must be 20 m above its launch position.
1
∆y = v iy t + ay t 2 gives
2
2 ⎛ 159 m ⎞
2 ⎛ 159 m ⎞
20 m = ( v i sin 35°) ⎜
⎟ − 4.90 m s ⎜ v ⎟ .
⎝ vi ⎠
⎝
⎠
i ( From which, v i = 42 m s .
(b) From above, t = 159 m 159 m
=
= 3.8 s
42 m s
vi 78 ) ) CHAPTER (c) 3 v x = v ix = ( 42 m s ) co s 35° = 34 m s ( ) v y = v iy + ay t = ( 42 m s ) sin 35° − 9.80 m s 2 ( 3.8 s ) = − 13 m s
2
2
v = vx + vy = 3.56 ( 34.1 m s )2 + ( −13.4 m s ) = 37 m s
2 We shall first find the initial velocity of the ball thrown vertically upward. At its
maximum height, v y = 0 and t = 1.50 s . Hence, v y = v iy + ay t gives ( ) 0 = v iy − 9.80 m s 2 (1.50 s ) , or v iy = 14.7 m s .
In order for the second ball to reach the same vertical height as the first, the second must
have the same initial vertical velocity. Thus, we find v i as vi = 3.57 v iy
sin 30.0° = 14.7 m s
= 29.4 m s .
0.500 1
The time of flight of the ball is given by ∆y = v iy t + ay t 2 , with ∆y = 0 , as
2
0 = ⎡( 20 m s ) sin 30°⎤ t +
⎣
⎦ ( ) 1
−9.80 m s 2 t 2 or t = 2.0 s.
2 The horizontal distance the football moves in this time is ∆x = v ix t = ⎡( 20 m s ) cos 30°⎤ ( 2.0 s ) = 35 m .
⎣
⎦
Therefore, the receiver must run a distance of (35 m – 20 m) = 15 m away from the
quarterback, in t h e d ir ect io n t h e b all w a s t h r o w n to catch the ball. He has a time of 2.0 s
to do this, so the required speed is
v= ∆x 15 m
=
= 7.5 m s
t
2.0 s 79 CHAPTER 3.58 3 The components of the initial velocity are v ix = v i co s 45° = vi
v
, and v iy = v i sin 45° = i .
2
2 The time for the ball to move 10.0 m horizontally is t = ∆x (10.0 m ) 2
=
.
v ix
vi At this time, the vertical displacement of the ball must be ∆y = ( 3.05 − 2.00) m = 1.05 m .
1
Thus, ∆y = v iy t + ay t 2 becomes
2 (10.0 m ) ( 2) ,
⎛ v ⎞ (10.0 m ) 2 1
1.05 m = ⎜ i ⎟
+ − 9.80 m s 2
⎝ 2⎠
2
vi
v i2 ( ) 2 which yields v i = 10.5 m s
3.59 Target Choose an origin where the projectile leaves the gun
and let the ycoordinates of the projectile and the target
at time t be labeled y p an d y T , respectively.
Then, ( ∆y ) p = y p − 0 = ( v 0 sin θo ) t − 2 t 2 , and h g ( ∆y )T = y T − h = 0 − 2 t 2
g or y T = h − v0 g2
t.
2 θ0
x0 The time when the projectile will have the same
x0
∆x
.
xcoordinate as the target is t =
=
v ix v o co s θ0
For a collision to occur, it is necessary that y p = yT at this time, or
⎛ g2
x0 ⎞ g 2
h
⎟ − 2 t = h − 2 t which reduces to tan θ o = x .
⎝ o co s θ0 ⎠
0 ( v 0 sin θo ) ⎜ v This requirement is satisfied provided that the gun is aimed at the initial location of the
target. Thus, a collision is guaranteed if the shooter aims the gun in this manner. 80 CHAPTER 3.60 3 (a) The components of the vectors are
Vector
d 1m
d 2m
d 1f
d 2f xcomponent (cm)
0
46.0
0
38.0 ycomponent (cm)
104
19.5
84.0
20.2 The sums d m = d 1m + d 2m and d f = d 1f + d 2f are computed as: dm = ( 0 + 46.0) 2 + (104 + 19.5) 2 = 132 cm an d θ = tan −1 ⎛
⎜
⎝ df = ( 0 + 38.0) 2 + ( 84.0 + 20.2) 2 = 111 cm an d θ = tan −1 ⎛
⎜
⎝ 104 + 19.5 ⎞
⎟ = 69.6°
0 + 46.0 ⎠ 84.0 + 20.2 ⎞
⎟ = 70.0°
0 + 38.0 ⎠ or d m = 132 cm a t 69.6° a n d d f = 111 cm a t 70.0° .
(b) To normalize, multiply each component in the above calculation by the appropriate
scale factor. The scale factor required for the components of d 1m a n d d 2m is
200 cm
sm =
= 1.11 , and the scale factor needed for components of d 1f a n d d 2f is
180 cm
200 cm
sf =
= 1.19 . After using these scale factors and recomputing the vector sums,
168 cm
the results are:
d m = 146 cm a t 69.6° a n d d f′ = 132 cm a t 70.0° .
′ The difference in the normalized vector sums is ∆d ′ = d m d f′ .
′
vector
dm
′
 d f′
∆d ′ xcomponent (cm)
50.9
–45.1
Σx = 5.74
Therefore, ∆d ′ = ycomponent (cm)
137
–124
Σy = 12.8 ( Σx ) 2 + ( Σy ) = ( 5.74) 2 + (12.8) cm = 14.0 cm , and
2 2 ⎛ Σy ⎞
⎛ 12 .8 ⎞
= t an −1 ⎜
= 65.8° , or ∆d ′ = 14.0 cm a t 65.8° .
⎝ Σx ⎟
⎠
⎝ 5.74 ⎟
⎠ θ = tan −1 ⎜ 81 CHAPTER 3.61 3 To achieve maximum range, the projectile should
be launched at 45° above the horizontal. In this
case, the initial velocity components are: vi 45° vi
.
2 v ix = v iy = R The time of flight may be found from v y = v iy − gt by recognizing that when the
projectile returns to the original level, v y = −v iy . Thus, the time of flight is t= −v iy − v iy
−g = 2v iy
g = 2 ⎛ vi ⎞ vi 2
. The maximum horizontal range is then
⎜ ⎟=
g ⎝ 2⎠
g 2
⎛ v ⎞ ⎛ v 2 ⎞ vi
R = v ix t = ⎜ i ⎟ ⎜ i
=
.
⎝ 2⎠ ⎝ g ⎟ g
⎠ (1) Now, consider throwing the projectile straight upward at speed v i . At maximum height,
v y = 0 , and the time required to reach this height is found from v y = v iy − gt as 0 = v i − gt
which yields t = vi
. Therefore, the maximum height the projectile will reach is
g ( ∆y )max = v y t = ⎛
⎜
⎝ 0 + v i ⎞ ⎛ v i ⎞ v i2
.
=
⎟
2 ⎠ ⎜ g ⎟ 2g
⎝⎠ Comparing this result with the maximum range found in equation (1) above reveals that
R
( ∆y )max = 2 provided the projectile is given the same initial speed in the two tosses.
If the boy takes a step when he makes the horizontal throw, he can likely give a higher
initial speed for that throw than for the vertical throw.
3.62 (a) At the top of the arc v y = 0 , and from v y = v iy − gt , we find the time to reach the top
of the arc to be t = v y − v iy
−g = 0 − v 0 s in θ o ⎛ v 0 ⎞
= ⎜ ⎟ sin θ o .
g
⎝ g⎠ The vertical height, h, reached in this time is found from
2
2
⎛ v y + v iy ⎞
⎛ 0 + v 0 sin θ0 ⎞ ⎛ v 0 sin θ0 ⎞ v 0 sin θ0
∆y = v y t = ⎜
t to be h = ⎜
=
.
⎟⎜
⎝
⎠⎝
2
g⎟
2g
⎝2⎟
⎠
⎠ 82 CHAPTER 3 (b) The total time of flight, t f , is double the time required to reach the top of the arc, or ⎛ 2v ⎞
t f = ⎜ 0 ⎟ sin θo . The horizontal range is given by
⎝g⎠
2
2
v 0 ( 2 sin θ0 cos θ0 ) v 0 sin ( 2θ0 )
⎛ 2v ⎞
R = v ix t f = ( v 0 co s θ0 ) ⎜ 0 ⎟ sin θo =
.
=
g
g
⎝g⎠ 3.63 The velocity of the boat relative to the shore is v bs = v bw + v w s , where v bw is the velocity
of the boat relative to the water and v ws is the velocity of the water relative to shore.
In order to cross the river (flowing parallel to the banks) in minimum time, the velocity
of the boat relative to the water must be perpendicular to the banks. That is, v bw must be
perpendicular to v ws . Hence, the velocity of the boat relative to the shore must be
2
2
v bs = v bw + v ws = (12 km h )2 + ( 5.0 km h )2 = 13 km h, ⎛ 12 km h ⎞
⎛v ⎞
= 67° to the direction of the current in the river
at θ = tan −1 ⎜ bw ⎟ = t an −1 ⎜
⎝ v ws ⎠
⎝ 5.0 k m h ⎟
⎠
(which is the same as the line of the riverbank).
The minimum time to cross the river is
t= w id t h o f r i v e r
1.5 k m ⎛ 60 m in ⎞
=
⎜
⎟ = 7.5 m in .
v bw
12 k m h ⎝ 1 h ⎠ During this time, the boat drifts downstream a distance of
3
⎛ 1 h ⎞ ⎛ 10 m ⎞
=6.3 × 102 m .
d = v ws t = ( 5.0 km h ) ( 7.5 m in ) ⎜
⎟⎜
⎝ 60 m in ⎠ ⎝ 1 km ⎟
⎠ 3.64 Taking upstream as positive, the velocity of the water relative to the ground is
v wg = − 0.500 m s . The velocity of the skater relative to shore is
+ 0.560 m
= + 0.700 m s while moving upstream, and v sg = v wg = − 0.500 m s while
0.800 s
drifting back downstream.
v sg = (a) At any time, v sg = v sw + v wg , or the velocity of the skater relative to the water is v sw = v sg − v wg .
(i) While going upstream, v sw = + 0.700 m s − ( − 0.500 m s ) = 1.20 m s .
83 CHAPTER 3 (ii) While drifting down stream, v sw = − 0.500 m s − ( − 0.500 m s ) = 0 .
(b) dsw = v sw t = (1.20 m s ) ( 0.800 s ) = 0.960 m
(c) The time to go upstream t up = 0.800 s and the time to drift back downstream is
t down = 0.560 m
= 1.12 s , giving the cycle time as 1.92 s.
0.500 m s Therefore, v sw = 3.65 dsw 0.960 m
=
= 0.500 m s .
t cycle
1.92 s The initial velocity components for the daredevil are v ix = v iy = 25.0 m s
vi
=
.
2
2 The time required to travel 50.0 m horizontally is t= ∆x ( 50.0 m ) 2
=
=2 2 s.
25.0 m s
v ix The vertical displacement of the daredevil at this time, and the proper height above the
level of the cannon to place the net, is
1
⎛ 25.0 m
∆y = v iy t + ay t 2 = ⎜
⎝
2
2 ( )( )( s⎞
2
⎟ 2 2 s − 9.80 m s 2 2 s
⎠ 84 ) 2 = 10.8 m . CHAPTER .0
m
/s At any time t, the horizontal and vertical
displacements of the projectile are: s = 15 ∆x = v ix t = ( v i co s 53.0°) t , and vi 3.66 3 53.0° 1
1
∆y = v iy t + ay t 2 = ( v i sin 53.0°) t − g t 2
2
2 20.0° ∆x Observe from the diagram that at the time
of impact, ∆y = ( ∆x ) tan 20.0° .
Therefore, at impact, we have ( v i sin 53.0°) t − g2
t = ⎡( v i co s 53.0°) t ⎤ t a n 20.0° .
⎣
⎦
2 Ignoring the t = 0 solution of this equation, this gives the time of impact as
t= = 2 vi
[ sin 53.0° − ( cos 53.0°) ta n 20.0°]
g
2 (15.0 m s ) [ sin 53.0° − ( co s 53.0°) t a n 20.0°]
9.80 m s 2 = 1.77 s At this time, the horizontal and vertical displacements are
∆x = ( v i co s 53.0°) t = (15.0 m s ) ( co s 53.0°) (1.77 s ) = 16.0 m , and ∆y = ( v i sin 53.0° ) t − 12
gt
2 = ( 15.0 m s ) ( sin 53.0° ) ( 1.77 s ) − ( 9.80 m s2 )
2 ( 1.77 s )2 = 5.83 m The straightline distance from the bottom of the ramp to the point of impact is then s= ( ∆x ) 2 + ( ∆y ) = (16.0 m ) 2 + ( 5.83 m ) 2 =
2 85 17.0 m . CHAPTER 3.67 3 (a) and (b)
Since the shot leaves the gun horizontally, the time it takes to reach the target is
∆x x
t=
= . The vertical displacement occurring in this time is
v ix v i
2 1
1 ⎛ x⎞
∆y = 0 − y = v iy t + ay t 2 = 0 − g ⎜ ⎟ , which gives the drop as
2
2 ⎝ vi ⎠
2 g
1 ⎛x⎞
y = g ⎜ ⎟ = A x 2 w ith A = 2 , w h er e v i is t h e m u z z le v elocity .
2 ⎝ vi ⎠
2 vi
(c) If x = 3.00 m, and y = 0.210 m, then A = vi = and 3.68 y
0.210 m
=
= 2.33 × 10−2 m 1
2
x
( 3.00 m ) 2 g
9.80 m s 2
=
= 14.5 m s .
2A
2 2 .33 × 10−2 m 1 ( ) Taking the positive x direction to be the direction of the first displacement, the
components of the three successive displacements are:
Displacement
1
2
3 xcomponent (m)
+10.0
0
–7.00
Σx = +3.00 m ycomponent (m)
0
–5.00
0
Σy = −5.00 m The resultant displacement is then R= ( Σx ) 2 + ( Σy ) = ( 3.00 m ) 2 + ( −5.00 m ) 2 = 5.83 m
2 ⎛ Σy ⎞
−1 ⎛ − 5.00 ⎞
= − 59.0°
⎟ = tan ⎜
⎝ Σx ⎠
⎝ 3.00 ⎟
⎠ θ = t a n −1 ⎜
or
3.69 R = 5.83 m a t 59.0° t o t h e r ig h t o f t h e o r ig in a l d ir ect io n . The components of the three displacements are:
Displacement
75.0 paces @ 240°
125 paces @ 135°
100 paces @ 160° xcomponent (paces)
–37.5
–88.4
–94.0
Σx = − 220 p a ces
86 ycomponent (paces)
–65.0
+88.4
+34.2
Σy = + 57.6 p a ce s CHAPTER 3 The resultant displacement is then R= ( Σx ) 2 + ( Σy ) = ( − 220 p aces ) + ( + 57.6 p aces ) = 227 p aces
2 2 2 ⎛ Σy ⎞
−1 ⎛ + 57.6 ⎞
= 165°
⎟ = t an ⎜
⎝ Σx ⎠
⎝ − 220 ⎟
⎠ θ = ta n −1 ⎜
or
3.70 R = 227 p a ces a t 165° fr o m t h e p o sit iv e x a xis . For the ball thrown at 45.0°, the time of flight is found from
g2
1
⎛v ⎞
∆y = v iy t + ay t 2 as 0 = ⎜ i ⎟ t1 − t1 ,
⎝ 2⎠
2
2 which has the single nonzero solution of t1 = vi 2
.
g 2
⎛ vi ⎞ ⎛ vi 2 ⎞ vi
=.
The horizontal range of this ball is R1 = v ix t1 = ⎜
⎝ 2⎟ ⎜ g ⎟ g
⎠⎝
⎠ Now consider the first arc in the motion of the second ball, started at angle θ with initial
1
speed v i . Applied to this arc, ∆y = v iy t + ay t 2 becomes
2
0 = ( v i sin θ ) t 21 − g2
t 21 ,
2 with nonzero solution t 21 = 2 v i s in θ
.
g Similarly, the time of flight for the second arc (started at angle θ with initial speed v i 2 )
of this ball’s motion is found to be
t 22 = 2 ( v i 2 ) s in θ
g = v i sin θ
.
g 87 CHAPTER 3 The horizontal displacement of the second ball during the first arc of its motion is ⎛ 2 v sin θ ⎞ v i2 ( 2 sin θ co s θ ) v i2 sin ( 2θ )
R21 = v ix t 21 = ( v i cos θ ) ⎜ i
.
=
⎟=
g
g
g
⎝
⎠
Similarly, the horizontal displacement during the second arc of this motion is R22 ( vi 2)2 sin ( 2θ ) = 1 v i2 sin ( 2θ ) .
=
g 4 g The total horizontal distance traveled in the two arcs is then
2
5 v i sin ( 2θ )
.
R2 = R 21 + R 22 =
4
g (a) Requiring that the two balls cover the same horizontal distance (i.e., requiring that
R 2 = R1 ) gives
2
5 v i sin ( 2θ ) v i2
.
=
4
g
g This reduces to sin ( 2θ ) = 4
which yields 2θ = 53.1° , so θ = 26.6° is the required
5
projection angle for the second ball. (b) The total time of flight for the second ball is
t 2 = t 21 + t 22 = 2 v i s i n θ v i s in θ 3 v i s i n θ
.
+
=
g
g
g Therefore, the ratio of the times of flight for the two balls is
t 2 ( 3 v i s in θ ) g
3
sin θ .
=
=
t1
2
vi 2 g ( ) With θ = 26.6° as found in (a), this becomes
t2
3
=
sin ( 26.6°) = 0.950 .
t1
2 88 CHAPTER 3.71 3 1
(a) Applying ∆y = v iy t + ay t 2 to the vertical motion of the first snowball gives
2
1
2
0 = ⎡( 25.0 m s ) sin 70.0°⎤ t1 + − 9.80 m s 2 t1 which has the nonzero solution of
⎣
⎦
2
2 ( 25.0 m s ) sin 70.0°
t1 =
= 4.79 s as the time of flight for this snowball.
9.80 m s 2 ( ) The horizontal displacement this snowball achieves is ∆x = v ix t1 = ⎡( 25.0 m s ) cos 70.0°⎤ ( 4.79 s ) = 41.0 m .
⎣
⎦
Now consider the second snowball, also given an initial speed of v i = 25.0 m s ,
1
thrown at angle θ, and is in the air for time t 2 . Applying ∆y = v iy t + ay t 2 to its
2
vertical motion yields
0 = ⎡( 25.0 m s ) sin θ ⎤ t 2 +
⎣
⎦ ( ) 1
2
−9.80 m s 2 t 2
2 which has a nonzero solution of
t2 = 2 ( 25.0 m s ) sin θ
9.80 m s 2 = ( 5.10 s ) sin θ . We require the horizontal range of this snowball be the same as that of the first ball,
namely ∆x = v ix t 2 = ⎡( 25.0 m s ) co s θ ⎤ ⎡( 5.10 s ) sin θ ⎤ = 41.0 m . This yields the equation
⎦
⎣
⎦⎣ sin θ cos θ = 41.0 m
= 0.321 .
( 25.0 m s ) ( 5.10 s ) From the trigonometric identity sin 2θ = 2 sin θ co s θ , this result becomes sin 2θ =2 ( 0.321) = 0.642 , so 2θ = 40.0°
and the required angle of projection for the second snowball is θ = 20.0° a b o v e t h e h o rizo n t a l . 89 CHAPTER 3 (b) From above, the time of flight for the first snowball is t1 = 4.79 s and that for the
second snowball is t 2 = ( 5.10 s ) sin θ = ( 5.10 s ) sin 20.0° = 1.74 s .
Thus, if they are to arrive simultaneously, the time delay between the first and
second snowballs should be
∆t = t1 − t 2 = 4.79 s − 1.74 s = 3.05 s . 90 CHAPTER 3 Answers to Even Numbered Conceptual Questions
2. The magnitudes add when A and B are in the same direction. The resultant will be zero
when the two vectors are equal in magnitude and opposite in direction. 4. The minimum sum for two vectors occurs when the two vectors are opposite in direction.
If they are unequal, their sum cannot add to zero. 6. The component of a vector can only be equal to or less than the vector itself. It can never
be greater than the vector. 8. The components of a vector will be equal in magnitude if the vector lies at a 45° angle with
the two axes along which the components lie. 10. They both start from rest in the downward direction and accelerate alike in the vertical
direction. Thus, they reach the ground with the same vertical speed. However, the ball
thrown horizontally had an initial horizontal component of velocity which is maintained
throughout the motion. Thus, the ball thrown horizontally moves with the greater speed. 12. The car can round a turn at a constant speed of 90 miles per hour. Its velocity will be
changing, however, because it is changing in direction. 14. The balls will be closest at the instant the second ball is projected. The first ball will always
be going faster than the second ball. There will be a one second time interval between their
collisions with the ground. The two move with the same acceleration in the vertical
direction. Thus, changing their horizontal velocity can never make them hit at the same
time. 16. Let v x and v y represent its original velocity components. We know that the vertical
component of velocity is zero at the top of the trajectory. Thus, 0 = v y − gt and the time at
the top of the trajectory is t = v y g .
(a) 2
vy
⎛ vy ⎞
x = v x ⎜ ⎟ and y =
2g
⎝ g⎠ (b) Its velocity is horizontal and equal to v x .
(c) Its acceleration is vertically downward, –g.
With air resistance, the answers to (a) and (b) would be smaller. As for (c) the
magnitude would be somewhat larger because the total acceleration would have a
component horizontally backward in addition to the vertical component of –g.
18. The equations of projectile motion are only valid for objects moving freely under the
influence of gravity. The only acceleration such an object has is the acceleration due to
gravity, g, directed vertically downward. Of the objects listed, only (a) and (d) meet this
requirement. 20. The passenger sees the ball go into the air and come back in the same way he would if he
were at rest on the Earth. An observer by the tracks would see the ball follow the path of a
projectile. If the train were accelerating, the ball would fall behind the position it would
reach in the absence of the acceleration.
91 CHAPTER 3 Answers to Even Numbered Problems
2. (a) 484 km 4. 83 m at 33° N of W 6. (a) 8. (a) 5.2 m at + 60°
(c) 3.0 m at + 150° 6.1 units at 113° (b) 18.1° N of W (b) 15 units at 23° (b) 3.0 m at 30°
(d) 5.20 m at 60° 10. 1.31 km north, 2.81 km east 12. 358 m at 2.00° S of E 14. 42.7 yards 16. 788 mi at 48.1° N of E 18. (a) 185 N at 77.8° (b) 185 N at 258° 20. (a) 74.6° N of E (b) 470 km 22. 2.65 ft (0.807 m) 24. 3.19 s, 36. m s at 60.1° below the horizontal
1 26. 0
4
2.8 m from base of table; vx = 5. m s, vy = −5. m s 28. x = 7. × 103 m , y = 1. × 103 m
23
68 30. (a) 32. 18.6 m 34. 9. m s
91 36. 61 s 38. (a) 10. m s at 8.53° E of N (b)
1 45.0 m 40. (a) 14.5° N of W 194 km h 42. 15.3 m 44. 7.87 N at 97.8° counterclockwise from a horizontal line to the right clears the bar by 0.85 m (b) (b) 4
falling, vy = − 13. m s 92 CHAPTER 3 (a) vix = 7. 9 m s, viy = 6. m s
9
02 (b) 1.85 m above the ground (c) 5.23 beyond the edge (d) 8. m s
97 50. (a) 0. m s
85 52. 14 m s 54. 30 m s 56. 29. m s
4 58. 10. m s
5 60. (a) 0
d m = 132 cm at69. ° , d f = 111 cm at70. °
6 (b) 0
d m = 146 cm at69. ° , d ′f = 132 cm at70. °
6
′ 46. (b) 2. m s
1 ∆d ′ = dm − d ′f = 14. cm at65. °
0
8
′
64. (a) 66. 17.0 m 68. 5.83 m at 59.0° to the right of the original direction 70. (a) 1. m s , 0
20 26.6° (b) (b) 0.960 m 0.950 93 (c) 0. 0 m s
50 ...
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This note was uploaded on 03/03/2010 for the course PHY P221 taught by Professor Dr.ha during the Spring '05 term at Indiana University South Bend.
 Spring '05
 Dr.Ha
 Physics

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