&
1. (a) The magnitude of r is

5.02 + ( −3.0) 2 + 2.02 = 6.2 m.

(b) A sketch is shown. The coordinate values are in meters.


2. Wherever the length unit is not specified (in this solution), the unit meter should be
understood.
&
(a) The position vector, according to Eq. 4-1, is r = ( − 5.0 m) ˆi + (8.0 m)jˆ .

&
(b) The magnitude is |r | =

x 2 + y 2 + z 2 = ( −5.0) 2 + (8.0) 2 + 0 2 = 9.4 m.

(c) Many calculators have polar ↔ rectangular conversion capabilities which make this
computation more efficient than what is shown below. Noting that the vector lies in the
xy plane, we are using Eq. 3-6:
§ 8.0 ·

θ = tan −1 ¨
¸ = − 58° or 122°
© −5.0 ¹
where we choose the latter possibility (122° measured counterclockwise from the +x
direction) since the signs of the components imply the vector is in the second quadrant.
(d) In the interest of saving space, we omit the sketch. The vector is 32° counterclockwise
from the +y direction, where the +y direction is assumed to be (as is standard) +90°
counterclockwise from +x, and the +z direction would therefore be “out of the paper.”

ˆ − (2.0iˆ − 3.0ˆj + 6.0 k)
ˆ = − 2.0 ˆi + 6.0 ˆj − 10 kˆ
ro = r − ∆r = (3.0ˆj − 4.0k)

where the understood unit is meters.


4. We choose a coordinate system with origin at the clock center and +x rightward
(towards the “3:00” position) and +y upward (towards “12:00”).
&
&
(a) In unit-vector notation, we have (in centimeters) r1 = 10i and r2 = − 10j. Thus, Eq.
4-2 gives
& &
&
∆ r = r2 − r1 = − 10iˆ − 10ˆj .

&
Thus, the magnitude is given by | ∆ r |= (− 10) 2 + (− 10) 2 = 14 cm.

(b) The angle is
§ − 10 ·
¸ = 45 ° or − 135 ° .
© − 10 ¹

θ = tan − 1 ¨

We choose − 135 ° since the desired angle is in the third quadrant. In terms of the
& &
&

¸ i = (40.0 km)i.
h
60
min/h
©
¹©
¹

The second displacement has a magnitude of 60.0
direction is 40° north of east. Therefore,

km
h

⋅

20.0 min
60 min/h

= 20.0 km, and its

&
∆r2 = 20.0 cos(40.0°) ˆi + 20.0 sin(40.0°) ˆj = 15.3 ˆi + 12.9 ˆj

in kilometers. And the third displacement is
&
km · § 50.0 min · ˆ
§
ˆ
∆r3 = − ¨ 60.0

© 2.90 ¹

θ = tan − 1 ¨
or 22.5° east of due north.


6. To emphasize the fact that the velocity is a function of time, we adopt the notation v(t)
for dx / dt.
(a) Eq. 4-10 leads to
v (t ) =

d
ˆ = (3.00 m/s)iˆ − (8.00t m/s) ˆj
(3.00tˆi − 4.00t 2 ˆj + 2.00k)
dt

&
ˆ m/s.
(b) Evaluating this result at t = 2.00 s produces v = (3.00iˆ − 16.0j)
&
(c) The speed at t = 2.00 s is v = |v | = (3.00) 2 + ( − 16.0) 2 = 16.3 m/s.
&
(d) And the angle of v at that moment is one of the possibilities

§ −16.0 ·
tan −1 ¨
¸ = − 79.4° or 101°
© 3.00 ¹

where we choose the first possibility (79.4° measured clockwise from the +x direction, or

(483) 2 +( − 966) 2 =1.08 × 103 km.

(b) The angle is given by
§ −966 ·
tan −1 ¨
¸ = − 63.4°.
© 483 ¹

We observe that the angle can be alternatively expressed as 63.4° south of east, or 26.6°
east of south.
&
(c) Dividing the magnitude of rAC by the total time (2.25 h) gives

483 ˆi − 966jˆ
&
vavg =
= 215iˆ − 429ˆj.
2.25

&
with a magnitude | vavg |= (215) 2 + (− 429) 2 =480 km/h.
&
(d) The direction of vavg is 26.6° east of south, same as in part (b). In magnitude-angle
&
notation, we would have vavg = (480 ∠ − 63.4 °).
&
(e) Assuming the AB trip was a straight one, and similarly for the BC trip, then | rAB | is the
&
distance traveled during the AB trip, and | rBC | is the distance traveled during the BC trip.
Since the average speed is the total distance divided by the total time, it equals

&
ro = 40i
&
r = 40j

and
and

&
vo = − 10j
&
v = 10i.

&
(a) Using Eq. 4-2, the displacement ∆ r is

& & &
∆ r = r − ro = − 40 ˆi+40 ˆj.
&
with a magnitude | ∆ r |= (− 40) 2 + (40)2 = 56.6 m.
&
(b) The direction of ∆ r is

§ ∆y ·
− 1 § 40 ·
¸ = tan ¨
¸ = − 45 ° or 135° .
© ∆x ¹
© − 40 ¹


¸ = 45 ° or − 135° .
© 0.333 ¹

θ = tan − 1 ¨

&
Since the desired angle is now in the first quadrant, we choose 45° , and aavg points
north of due east.


11. In parts (b) and (c), we use Eq. 4-10 and Eq. 4-16. For part (d), we find the direction
of the velocity computed in part (b), since that represents the asked-for tangent line.
(a) Plugging into the given expression, we obtain
&
r

t = 2.00

= [2.00(8) − 5.00(2)]iˆ + [6.00 − 7.00(16)] ˆj = 6.00 ˆi − 106 ˆj

in meters.
(b) Taking the derivative of the given expression produces
&
v (t ) = (6.00t 2 − 5.00) ˆi − 28.0t 3 ˆj

where we have written v(t) to emphasize its dependence on time. This becomes, at
&
t = 2.00 s, v = (19.0 ˆi − 224 ˆj) m/s.
&
(c) Differentiating the v ( t ) found above, with respect to t produces 12.0t ˆi − 84.0t 2 ˆj,

Thus, the magnitude of v is | v |= (11.7) 2 + (10.7)2 = 15.8 m/s.
&
(b) The angle of v , measured from +x, is

§ 10.7 ·
tan −1 ¨
¸ = 42.6°.
© 11.7 ¹


13. We find t by applying Eq. 2-11 to motion along the y axis (with vy = 0 characterizing
y = ymax ): 0 = (12 m/s) + (−2.0 m/s2)t Ÿ t = 6.0 s. Then, Eq. 2-11 applies to motion
along the x axis to determine the answer: vx = (8.0 m/s) + (4.0 m/s2)(6.0 s) = 32 m/s.
Therefore, the velocity of the cart, when it reaches y = ymax , is (32 m/s)i^.


14. We make use of Eq. 4-16.
(a) The acceleration as a function of time is
&
& dv d
a=
=
dt dt

( ( 6.0t − 4.0t ) ˆi + 8.0 ˆj) = ( 6.0 − 8.0t ) ˆi
2

in SI units. Specifically, we find the acceleration vector at t = 3.0 s to be
( 6.0 − 8.0(3.0) ) ˆi = (−18 m/s2 )i.ˆ


rearranging 4.0t 2 − 6.0t ± 6.0 = 0
using quadratic formula t =

6.0 ± 36 − 4 ( 4.0 )( ±6.0 )
2 ( 8.0 )

where the requirement of a real positive result leads to the unique answer: t = 2.2 s.


15. Constant acceleration in both directions (x and y) allows us to use Table 2-1 for the
motion along each direction. This can be handled individually (for ∆x and ∆y) or together
with the unit-vector notation (for ∆r). Where units are not shown, SI units are to be
understood.
& &
&
&
(a) The velocity of the particle at any time t is given by v = v0 + at , where v0 is the
&
initial velocity and a is the (constant) acceleration. The x component is vx = v0x + axt =
3.00 – 1.00t, and the y component is vy = v0y + ayt = –0.500t since v0y = 0. When the
particle reaches its maximum x coordinate at t = tm, we must have vx = 0. Therefore, 3.00
– 1.00tm = 0 or tm = 3.00 s. The y component of the velocity at this time is

vy = 0 – 0.500(3.00) = –1.50 m/s;
&
this is the only nonzero component of v at tm.

(b) Since it started at the origin, the coordinates of the particle at any time t are given by
& &
&


g

b

g

1
0.40 cosθ t 2 .
2

Second, the x motions of A and B must coincide:
vt =

1 2
1
a x t Ÿ 3.0t = 0.40 sin θ t 2 .
2
2

We eliminate a factor of t in the last relationship and formally solve for time:
t=

3.0
.
0.20 sin θ

This is then plugged into the previous equation to produce
1
30 = ( 0.40 cos θ )


θ = cos−1

FG 1 IJ = 60° .
H 2K

=

1
2


17. (a) From Eq. 4-22 (with θ0 = 0), the time of flight is
t=

2h
2(45.0)
=
= 3.03 s.
g
9.80

(b) The horizontal distance traveled is given by Eq. 4-21:
∆x = v0 t = ( 250)( 3.03) = 758 m.

(c) And from Eq. 4-23, we find
v y = gt = ( 9.80)( 3.03) = 29.7 m / s.


18. We use Eq. 4-26

problem) to first find the initial y velocity, that is how we will proceed. Using Eq. 2-16,
we have
v12 y = v02y − 2 g ∆y Ÿ (6.1)2 = v02 y − 2(9.8)(9.1)
which yields v0 y = 14.7 m/s. Knowing that v2 y must equal 0, we use Eq. 2-16 again but
now with ∆y = h for the maximum height:
v22 y = v02 y − 2 gh Ÿ

0 = (14.7) 2 − 2(9.8)h

which yields h = 11 m.
(b) Recalling the derivation of Eq. 4-26, but using v0 y for v0 sin θ0 and v0x for v0 cos θ0,
we have

0 = v0 y t −

1 2
gt
2

R = v0 x t
which leads to R = 2v0 x v0 y / g . Noting that v0x = v1x = 7.6 m/s, we plug in values and
obtain R = 2(7.6)(14.7)/9.8 = 23 m.
(c) Since v3x = v1x = 7.6 m/s and v3y = – v0 y = –14.7 m/s, we have
v3 = v32 x + v32 y = (7.6) 2 + (−14.7) 2 = 17 m/s.
&
(d) The angle (measured from horizontal) for v3 is one of these possibilities:

§ −14.7 ·
tan −1 ¨
¸ = −63° or 117°


21. We adopt the positive direction choices used in the textbook so that equations such as
Eq. 4-22 are directly applicable. The initial velocity is horizontal so that v 0 y = 0 and
v0 x = v0 = 10 m s.
(a) With the origin at the initial point (where the dart leaves the thrower’s hand), the y
coordinate of the dart is given by y = − 21 gt 2 , so that with y = –PQ we have
PQ =

1
2

. g
b9.8gb019

2

= 018
. m.

(b) From x = v0t we obtain x = (10)(0.19) = 1.9 m.


22. (a) Using the same coordinate system assumed in Eq. 4-22, we solve for y = h:
h = y0 + v0 sin θ 0t −

1 2
gt
2

which yields h = 51.8 m for y0 = 0, v0 = 42.0 m/s, θ0 = 60.0° and t = 5.50 s.

at the moment of release: v0 = 290 km/h, which we convert to SI units: (290)(1000/3600)
= 80.6 m/s.
(a) We use Eq. 4-12 to solve for the time:
∆x = (v0 cos θ 0 ) t

Ÿ t=

700
= 10.0 s.
(80.6) cos (−30.0°)

(b) And we use Eq. 4-22 to solve for the initial height y0:
1
1
y − y0 = (v0 sin θ 0 ) t − gt 2 Ÿ 0 − y0 = (−40.3)(10.0) − (9.80)(10.0) 2
2
2

which yields y0 = 897 m.


24. We adopt the positive direction choices used in the textbook so that equations such as
Eq. 4-22 are directly applicable. The coordinate origin is throwing point (the stone’s
initial position). The x component of its initial velocity is given by v0 x = v0 cosθ 0 and the
y component is given by v 0 y = v 0 sin θ 0 , where v0 = 20 m/s is the initial speed and θ 0 =
40.0° is the launch angle.
(a) At t = 1.10 s, its x coordinate is

b



1
9.80m/s 2 ) (1.80s 2 ) = 7.26m.
(
2

(e) The stone hits the ground earlier than t = 5.0 s. To find the time when it hits the
ground solve y = v0 t sin θ 0 − 21 gt 2 = 0 for t. We find
t=

b

g

2 20.0 m / s
2 v0
sin 40° = 2.62 s.
sin θ 0 =
g
9.8 m / s2

Its x coordinate on landing is

b

gb

g

x = v0t cos θ 0 = 20.0 m / s 2.62 s cos 40° = 40.2 m