Content check — University Physics 1, Chapter 4: Motion in Two and Three Dimensions

Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 4: Motion in Two and Three Dimensions

57 problems · 11 figures

Check Your Understanding 4.1 §4.1 source ↗

The position function of a particle is

\vec{r} (t) =

3.0 t^{3} \hat{i} +

4.0 \hat{j} .

(a) What is the instantaneous velocity at t = 3 s? (b) Is the average velocity between 2 s and 4 s equal to the instantaneous velocity at t = 3 s?

Answer

(a) Taking the derivative with respect to time of the position function, we have

\vec{v} (t) =

9.0 t^{2} \hat{i} and \vec{v} (3.0s) =

81.0 \hat{i} m/s .

(b) Since the velocity function is nonlinear, we suspect the average velocity is not equal to the instantaneous velocity. We check this and find

{\vec{v}}_{avg} =

\frac{\vec{r} (t_{2}) - \vec{r} (t_{1})}{t_{2} - t_{1}} =

\frac{\vec{r} (4.0 s) - \vec{r} (2.0 s)}{4.0 s - 2.0 s} =

\frac{[(192 - 24) \hat{i} + (4 - 4) \hat{i}] m}{2.0 s} =

84 \hat{j} m/s,

which is different from

\vec{v} (3.0s) =

81.0 \hat{i} m/s .

Check Your Understanding 4.2 §4.2 source ↗

Suppose the acceleration function has the form

\vec{a} (t) =

a \hat{i} +

b \hat{j} +

c \hat{k} m/ s^{2},

where a, b, and c are constants. What can be said about the functional form of the velocity function?

Answer

The acceleration vector is constant and doesn’t change with time. If a, b, and c are not zero, then the velocity function must be linear in time. We have

\vec{v} (t) =

\int \vec{a} d t = \int (a \hat{i} + b \hat{j} + c \hat{k}) d t = (a \hat{i} + b \hat{j} + c \hat{k}) t m/s,

since taking the derivative of the velocity function produces

\vec{a} (t) .

If any of the components of the acceleration are zero, then that component of the velocity would be a constant.

Check Your Understanding 4.3 §4.3 source ↗

A rock is thrown horizontally off a cliff

100.0 m

high with a velocity of 15.0 m/s. (a) Define the origin of the coordinate system. (b) Which equation describes the horizontal motion? (c) Which equations describe the vertical motion? (d) What is the rock’s velocity at the point of impact?

Answer

(a) Choose the top of the cliff where the rock is thrown from the origin of the coordinate system. Although it is arbitrary, we typically choose time t = 0 to correspond to the origin. (b) The equation that describes the horizontal motion is

x = x_{0} + v_{x} t .

With

x_{0} = 0,

this equation becomes

x = v_{x} t .

y_{0} =

0 and v_{0 y} =

0,

these equations simplify greatly to become

y =

\frac{1}{2} (v_{0 y} + v_{y}) t =

\frac{1}{2} v_{y} t,

v_{y} = − g t,

y = - \frac{1}{2} g t^{2},

and

v_{y}^{2} = −2 g y .

(d) We use the kinematic equations to find the x and y components of the velocity at the point of impact. Using

v_{y}^{2} = −2 g y

and noting the point of impact is −100.0 m, we find the y component of the velocity at impact is

v_{y} = 44.3 m / s .

We are given the x component,

v_{x} = 15.0 m / s,

so we can calculate the total velocity at impact: v = 46.8 m/s and

θ = 71.3 °

below the horizontal.

Check Your Understanding 4.4 §4.3 source ↗

If the two golf shots in Example 4.9 were launched at the same speed, which shot would have the greatest range?

Answer

The golf shot at

30 ° .

Check Your Understanding 4.5 §4.4 source ↗

A flywheel has a radius of 20.0 cm. What is the speed of a point on the edge of the flywheel if it experiences a centripetal acceleration of

900.0 {cm / s}^{2} ?

Answer

134.0 cm/s

Check Your Understanding 4.6 §4.5 source ↗

A boat heads due north, moving at 4.5 m/s relative to the water of a river that is running east at 3.0 m/s. What is the velocity of the boat with respect to Earth?

Answer

Labeling subscripts for the vector equation, we have B = boat, R = river, and E = Earth. The vector equation becomes ${\vec{v}}_{BE} =$ ${\vec{v}}_{BR} +$ ${\vec{v}}_{RE} .$ We have right triangle geometry shown in Figure 04_05_BoatRiv_img. Solving for ${\vec{v}}_{BE}$ , we have
$v_{BE} =$ $\sqrt{v_{BR}^{2} + v_{RE}^{2}} =$ $\sqrt{{4.5}^{2} + {3.0}^{2}}$
$v_{BE} =$ $5.4 m / s, θ =$ ${tan}^{−1} (\frac{3.0}{4.5}) =$ $33.7 ° .$

Vectors V sub B W, V sub W E and V sub B E form a right triangle. A boat is shown at the vertex where the tails of V sub B W and V sub B E meet. Vector V sub B W points up. V sub W E points to the right. V sub B E points up and right, at an angle to the vertical. V sub B E is the vector sum of v sub B W and V sub W E.

4-1 §4.1 source ↗

What form does the trajectory of a particle have if the distance from any point A to point B is equal to the magnitude of the displacement from A to B?

Answer

straight line

4-3 §4.1 source ↗

If the instantaneous velocity is zero, what can be said about the slope of the position function?

Answer

The slope must be zero because the velocity vector is tangent to the graph of the position function.

4-5 §4.2 source ↗

If an object has a constant x-component of the velocity and suddenly experiences an acceleration in the y direction, does the x-component of its velocity change?

Answer

No, motions in perpendicular directions are independent.

4-7 §4.3 source ↗

Answer the following questions for projectile motion on level ground assuming negligible air resistance, with the initial angle being neither

0 °

nor

90 ° :

(a) Is the velocity ever zero? (b) When is the velocity a minimum? A maximum? (c) Can the velocity ever be the same as the initial velocity at a time other than at t = 0? (d) Can the speed ever be the same as the initial speed at a time other than at t = 0?

Answer

a. no; b. minimum at apex of trajectory and maximum at launch and impact; c. no, velocity is a vector; d. yes, where it lands

4-9 §4.3 source ↗

A dime is placed at the edge of a table so it hangs over slightly. A quarter is slid horizontally on the table surface perpendicular to the edge and hits the dime head on. Which coin hits the ground first?

Answer

They both hit the ground at the same time.

4-11 §4.4 source ↗

Can tangential acceleration change the speed of a particle undergoing circular motion?

Answer

yes

4-13 §4.5 source ↗

A basketball player dribbling down the court usually keeps his eyes fixed on the players around him. He is moving fast. Why doesn’t he need to keep his eyes on the ball?

Answer

If he is going to pass the ball to another player, he needs to keep his eyes on the reference frame in which the other players on the team are located.

4-15 §4.5 source ↗

The hat of a jogger running at constant velocity falls off the back of his head. Draw a sketch showing the path of the hat in the jogger’s frame of reference. Draw its path as viewed by a stationary observer. Neglect air resistance.

Answer

Figure a: a hat’s trajectory is straight down. Figure b: a hat’s trajectory is parabolic, curving down and to the left.

4-17 §4.1 source ↗

The coordinates of a particle in a rectangular coordinate system are (1.0, –4.0, 6.0). What is the position vector of the particle?

Answer

\vec{r} =

1.0 \hat{i} -

4.0 \hat{j} +

6.0 \hat{k}

4-19 §4.1 source ↗

The 18th hole at Pebble Beach Golf Course is a dogleg to the left of length 496.0 m. The fairway off the tee is taken to be the x direction. A golfer hits his tee shot a distance of 300.0 m, corresponding to a displacement

Δ {\vec{r}}_{1} =

300.0 m \hat{i},

and hits his second shot 189.0 m with a displacement

Δ {\vec{r}}_{2} =

172.0 m \hat{i} +

80.3 m \hat{j} .

What is the final displacement of the golf ball from where it started?

Answer

Δ {\vec{r}}_{Total} =

472.0 m \hat{i} +

80.3 m \hat{j}

4-21 §4.1 source ↗

A cyclist rides 5.0 km due east, then 10.0 km

20 °

west of north. From this point she rides 8.0 km due west. What is the final displacement from where the cyclist started?

Answer

Sum of displacements =

−6.4 km \hat{i} +

9.4 km \hat{j}

4-23 §4.1 source ↗

The position of a particle is

\vec{r} (t) =

4.0 t^{2} \hat{i} -

3.0 \hat{j} +

2.0 t^{3} \hat{k} m .

(a) What is the velocity of the particle at 0 s and at

1.0

s? (b) What is the average velocity between 0 s and

1.0

Answer

\vec{v} (t) =

8.0 t \hat{i} +

6.0 t^{2} \hat{k},

\vec{v} (0) =

0,

\vec{v} (1.0) =

8.0 \hat{i} +

6.0 \hat{k} m/s

,
b.

{\vec{v}}_{avg} =

4.0 ​ \hat{i} +

2.0 \hat{k} m/s

4-25 §4.1 source ↗

The F-35B Lighting II is a short-takeoff and vertical landing fighter jet. If it does a vertical takeoff to 20.00-m height above the ground and then follows a flight path angled at

30 °

with respect to the ground for 20.00 km, what is the final displacement?

Answer

Δ {\vec{r}}_{1} =

20.00 m \hat{j},

Δ {\vec{r}}_{2} =

(2.000 \times 10^{4} m) (cos 30 ° \hat{i} + sin 30 ° \hat{j})

Δ \vec{r} =

1.700 \times 10^{4} m \hat{i} +

1.002 \times 10^{4} m \hat{j}

4-27 §4.2 source ↗

A particle’s acceleration is

(4.0 \hat{i} + 3.0 \hat{j}) m/ s^{2} .

At t = 0, its position and velocity are zero. (a) What are the particle’s position and velocity as functions of time? (b) Find the equation of the path of the particle. Draw the x- and y-axes and sketch the trajectory of the particle.

Answer

a. $\vec{v} (t) =$ $(4.0 t \hat{i} + 3.0 t \hat{j}) m/s,$ $\vec{r} (t) =$ $(2.0 t^{2} \hat{i} + \frac{3}{2} t^{2} \hat{j}) m$ ,
b. $x (t) =$ $2.0 t^{2} m, y (t) =$ $\frac{3}{2} t^{2} m, t^{2} =$ $\frac{x}{2} \Rightarrow y =$ $\frac{3}{4} x$

A graph of the linear function y equals 3 quarters x. The graph is a straight, positive slope line through the origin.

4-29 §4.2 source ↗

The position of a particle for t > 0 is given by

\vec{r} (t) =

(3.0 t^{2} \hat{i} - 7.0 t^{3} \hat{j} - 5.0 t^{−2} \hat{k}) m .

(a) What is the velocity as a function of time? (b) What is the acceleration as a function of time? (c) What is the particle’s velocity at t = 2.0 s? (d) What is its speed at t = 1.0 s and t = 3.0 s? (e) What is the average velocity between t = 1.0 s and t = 2.0 s?

Answer

\vec{v} (t) =

(6.0 t \hat{i} - 21.0 t^{2} \hat{j} + 10.0 t^{−3} \hat{k}) m/s

\vec{a} (t) =

(6.0 \hat{i} - 42.0 t \hat{j} - 30 t^{−4} \hat{k}) m/ s^{2}

\vec{v} (2.0 s) =

(12.0 \hat{i} - 84.0 \hat{j} + 1.25 \hat{k}) m/s

\vec{v} (1.0 s) =

6.0 \hat{i} -

21.0 \hat{j} +

10.0 \hat{k} m/s,

| \vec{v} (1.0 s) | =

24.0 m/s

\vec{v} (3.0 s) =

18.0 \hat{i} -

189.0 \hat{j} +

0.37 \hat{k} m/s,

| \vec{v} (3.0 s) | =

190 m/s

\vec{r} (t) =

(3.0 t^{2} \hat{i} - 7.0 t^{3} \hat{j} - 5.0 t^{−2} \hat{k}) m

{\vec{v}}_{avg} =

9.0 \hat{i} -

49.0 \hat{j} +

3.75 \hat{k} m/s

4-31 §4.2 source ↗

A particle has a position function

\vec{r} (t) =

cos (1.0 t) \hat{i} +

sin (1.0 t) \hat{j} +

t \hat{k},

where the arguments of the cosine and sine functions are in radians. (a) What is the velocity vector? (b) What is the acceleration vector?

Answer

\vec{v} (t) =

−sin (1.0 t) \hat{i} +

cos (1.0 t) \hat{j} +

\hat{k}

, b.

\vec{a} (t) =

−cos (1.0 t) \hat{i} -

sin (1.0 t) \hat{j}

4-33 §4.3 source ↗

A bullet is shot horizontally from shoulder height (1.5 m) with an initial speed 200 m/s. (a) How much time elapses before the bullet hits the ground? (b) How far does the bullet travel horizontally?

Answer

t = 0.55 s

, b.

x = 110 m

4-35 §4.3 source ↗

A dart is thrown horizontally at a speed of 10 m/s at the bull’s-eye of a dartboard 2.4 m away, as in the following figure. (a) How far below the intended target does the dart hit? (b) What does your answer tell you about how proficient dart players throw their darts?

An illustration of a person throwing a dart. The dart is released horizontally a distance of 2.4 meters from the dart board, level with the bulls eye of the dart board, with a speed of 10 meters per second.

Answer

a. $t =$ $0.24 s, d =$ $0.28 m$ , b. They aim high.

4-37 §4.3 source ↗

Suppose the airplane in the preceding problem fires a projectile horizontally in its direction of motion at a speed of 300 m/s relative to the plane. (a) How far in front of the release point does the projectile hit the ground? (b) What is its speed when it hits the ground?

Answer

a.,

t =

12.8 s, x =

5619 m

v_{y} =

125.0 m / s, v_{x} =

439.0 m / s, | \vec{v} | =

456.0 m / s

4-39 §4.3 source ↗

A projectile is launched at an angle of

30 °

and lands 20 s later at the same height as it was launched. (a) What is the initial speed of the projectile? (b) What is the maximum altitude? (c) What is the range? (d) Calculate the displacement from the point of launch to the position on its trajectory at 15 s.

Answer

v_{y} =

v_{0 y} -

g t,

t =

10 s, v_{y} =

0,

v_{0 y} =

98.0 m / s,

v_{0} =

196.0 m / s

,
b.

h = 490.0 m,

v_{0 x} =

169.7 m / s, x =

3394.0 m,

\begin{matrix} x = 169.7 m/s (15.0 s) = 2550 m \\ y = (98.0 m/s) (15.0 s) - 4.9 (15.0 s)^{2} = 368 m \\ \overset{r}{s} = 2550 m \hat{i} + 368 m \hat{j} \end{matrix}

4-41 §4.3 source ↗

At a particular instant, a hot air balloon is 100 m in the air and descending at a constant speed of 2.0 m/s. At this exact instant, a girl throws a ball horizontally, relative to herself, with an initial speed of 20 m/s. When she lands, where will she find the ball? Ignore air resistance.

Answer

−100 m =

(−2.0 m / s) t -

(4.9 m / s^{2}) t^{2},

t = 4.3 s,

x = 86.0 m

4-43 §4.3 source ↗

An athlete can jump a distance of 8.0 m in the broad jump. What is the maximum distance the athlete can jump on the Moon, where the gravitational acceleration is one-sixth that of Earth?

Answer

R_{M o o n} = 48 m

4-45 §4.3 source ↗

A rock is thrown off a cliff at an angle of

53 °

with respect to the horizontal. The cliff is 100 m high. The initial speed of the rock is 30 m/s. (a) How high above the edge of the cliff does the rock rise? (b) How far has it moved horizontally when it is at maximum altitude? (c) How long after the release does it hit the ground? (d) What is the range of the rock? (e) What are the horizontal and vertical positions of the rock relative to the edge of the cliff at t = 2.0 s, t = 4.0 s, and t = 6.0 s?

Answer

v_{0 y} = 24 m / s

v_{y}^{2} =

v_{0 y}^{2} -

2 g y \Rightarrow h =

29.3 m

,
b.

t =

2.4 s v_{0 x} =

18 m/s x =

43.2 m

,
c.

y = −100 m y_{0} = 0

y -

y_{0} =

v_{0 y} t -

\frac{1}{2} g t^{2} -

100 =

24 t -

4.9 t^{2}

\Rightarrow t = 7.58 s

,
d.

x = 136.44 m

,
e.

t =

2.0 s y =

28.4 m x =

36 m

t =

4.0 s y =

17.6 m x =

72 m

t =

6.0 s y =

−32.4 m x =

108 m

4-47 §4.3 source ↗

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of

40 °

with an initial speed of 20 m/s, how close to the green does she come?

Answer

v_{0 y} =

12.9 m / s y -

y_{0} =

v_{0 y} t -

\frac{1}{2} g t^{2} -

20.0 =

12.9 t -

4.9 t^{2}

t =

3.7 s v_{0 x} =

15.3 m / s \Rightarrow x =

56.7 m

So the golfer’s shot lands 13.3 m short of the green.

4-49 §4.3 source ↗

An astronaut on Mars kicks a soccer ball at an angle of

45 °

with an initial velocity of 15 m/s. If the acceleration of gravity on Mars is

3.7 m/ s^{2}

, (a) what is the range of the soccer kick on a flat surface? (b) What would be the range of the same kick on the Moon, where gravity is one-sixth that of Earth?

Answer

R = 60.8 m

,
b.

R = 137.8 m

4-51 §4.3 source ↗

MIT’s robot cheetah can jump over obstacles 46 cm high and has speed of 12.0 km/h. (a) If the robot launches itself at an angle of

60 °

at this speed, what is its maximum height? (b) What would the launch angle have to be to reach a height of 46 cm?

Answer

v_{y}^{2} =

v_{0 y}^{2} -

2 g y \Rightarrow y =

2.9 m / s

y = 3.3 m / s

y =

\frac{v_{0 y}^{2}}{2 g} =

\frac{{(v_{0} sin θ)}^{2}}{2 g} \Rightarrow sin θ =

0.91 \Rightarrow θ =

65.5 °

4-53 §4.3 source ↗

Drew Brees of the New Orleans Saints can throw a football 23.0 m/s (50 mph). If he angles the throw at

10 °

from the horizontal, what distance does it go if it is to be caught at the same elevation as it was thrown?

Answer

R = 18.5 m

4-55 §4.3 source ↗

A soccer goal is 2.44 m high. A player kicks the ball at a distance 10 m from the goal at an angle of

25 ° .

The ball hits the crossbar at the top of the goal. What is the initial speed of the soccer ball?

Answer

y =

(tan θ_{0}) x -

[\frac{g}{2 {(v_{0} cos θ_{0})}^{2}}] x^{2} \Rightarrow v_{0} =

16.4 m / s

4-57 §4.3 source ↗

In 1999, Robbie Knievel was the first to jump the Grand Canyon on a motorcycle. At a narrow part of the canyon (69.0 m wide) and traveling 35.8 m/s off the takeoff ramp, he reached the other side. What was his launch angle?

Answer

R =

\frac{v_{0}^{2} sin 2 θ_{0}}{g} \Rightarrow θ_{0} =

15.9 °

4-59 §4.3 source ↗

Aaron Rodgers throws a football at 20.0 m/s to his wide receiver, who is running straight down the field at 9.4 m/s. If Aaron throws the football when the wide receiver is 10.0 m in front of him, (a) at what angle does Aaron have to launch the ball so the ball will be at the same height as the receiver when the receiver makes it to 20.0 m in front of Aaron? (b) Will the receiver be able to catch the ball?

Answer

(a) It takes the wide receiver 1.1 s to cover the last 10 m of his run.

T_{tof} =

\frac{2 (v_{0} sin θ)}{g} \Rightarrow sin θ =

0.27 \Rightarrow θ =

15.6 °

(b)

x =

\frac{v_{0}^{2} \sin (2 θ)}{g} =

21 m

Therefore, the ball will be overthrown, and the receiver will not be able to catch it.

4-61 §4.4 source ↗

A particle travels in a circle of radius 10 m at a constant speed of 20 m/s. What is the magnitude of the acceleration?

Answer

a_{C} = 40 m / s^{2}

4-63 §4.4 source ↗

A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has an 8.00-m radius, at how many revolutions per minute are the riders subjected to a centripetal acceleration equal to that of gravity?

Answer

a_{C} =

\frac{v^{2}}{r} \Rightarrow v^{2} =

r a_{C} =

78.4,

v =

8.85 m / s

T = 5.68 s,

which is

0.176 rev / s =

10.6 rev / min

4-65 §4.4 source ↗

What is the acceleration of Venus toward the Sun, assuming a circular orbit?

Answer

Venus is 108.2 million km from the Sun and has an orbital period of 0.6152 y.

r =

1.082 \times 10^{11} m T =

1.94 \times 10^{7} s

v =

3.5 \times 10^{4} m/s, a_{C} =

1.135 \times 10^{−2} {m/s}^{2}

4-67 §4.4 source ↗

A fan is rotating at a constant 360.0 rev/min. What is the magnitude of the acceleration of a point on one of its blades 10.0 cm from the axis of rotation?

Answer

360 rev / min =

6 rev / s

v = 3.8 m / s

a_{C} = 144. m / s^{2}

4-69 §4.5 source ↗

The coordinate axes of the reference frame

S^{'}

remain parallel to those of

S

, as

S^{'}

moves away from

S

at a constant velocity

{\vec{v}}_{S^{'}}^{S} =

(4.0 \hat{i} + 3.0 \hat{j} + 5.0 \hat{k}) m / s .

(a) If at time t = 0 the origins coincide, what is the position of the origin

O^{'}

in the S frame as a function of time? (b) How is particle position for

\vec{r} (t)

and

{\vec{r}}^{'} (t),

as measured in S and

S^{'},

respectively, related? (c) What is the relationship between particle velocities

\vec{v} (t) and {\vec{v}}^{'} (t) ?

(d) How are accelerations

\vec{a} (t) and {\vec{a}}^{'} (t)

related?

Answer

O^{'} (t) =

(4.0 \hat{i} + 3.0 \hat{j} + 5.0 \hat{k}) t m

,
b.

{\vec{r}}_{P S} =

{\vec{r}}_{P S^{'}} +

{\vec{r}}_{S^{'} S},

\vec{r} (t) =

{\vec{r}}^{'} (t) +

(4.0 \hat{i} + 3.0 \hat{j} + 5.0 \hat{k}) t m

,
c.

\vec{v} (t) =

{\vec{v}}^{'} (t) +

(4.0 \hat{i} + 3.0 \hat{j} + 5.0 \hat{k}) m/s

, d. The accelerations are the same.

4-71 §4.5 source ↗

The velocity of a particle in reference frame A is

(2.0 \hat{i} + 3.0 \hat{j}) m / s .

The velocity of reference frame A with respect to reference frame B is

4.0 \hat{k} m / s,

and the velocity of reference frame B with respect to C is

2.0 \hat{j} m / s .

What is the velocity of the particle in reference frame C?

Answer

{\vec{v}}_{P C} =

(2.0 \hat{i} + 5.0 \hat{j} + 4.0 \hat{k}) m / s

4-73 §4.5 source ↗

A seagull can fly at a velocity of 9.00 m/s in still air. (a) If it takes the bird 20.0 min to travel 6.00 km straight into an oncoming wind, what is the velocity of the wind? (b) If the bird turns around and flies with the wind, how long will it take the bird to return 6.00 km?

Answer

a. A = air, S = seagull, G = ground

{\vec{v}}_{SA} = 9.0 m / s

velocity of seagull with respect to still air

{\vec{v}}_{AG} =

? {\vec{v}}_{SG} =

5 m / s

{\vec{v}}_{SG} =

{\vec{v}}_{SA} +

{\vec{v}}_{AG} \Rightarrow {\vec{v}}_{AG} =

{\vec{v}}_{SG} -

{\vec{v}}_{SA}

{\vec{v}}_{AG} =

−4.0 m / s

{\vec{v}}_{SG} =

{\vec{v}}_{SA} +

{\vec{v}}_{AG} \Rightarrow {\vec{v}}_{SG} =

−13.0 m / s

\frac{−6000 m}{−13.0 m / s} =

7 min 42 s

4-75 §4.5 source ↗

A boat can be rowed at 8.0 km/h in still water. (a) How much time is required to row 1.5 km downstream in a river moving 3.0 km/h relative to the shore? (b) How much time is required for the return trip? (c) In what direction must the boat be aimed to row straight across the river? (d) Suppose the river is 0.8 km wide. What is the velocity of the boat with respect to Earth and how much time is required to get to the opposite shore? (e) Suppose, instead, the boat is aimed straight across the river. How much time is required to get across and how far downstream is the boat when it reaches the opposite shore?

Answer

Take the positive direction to be the same direction that the river is flowing, which is east. S = shore/Earth, W = water, and B = boat.
a. ${\vec{v}}_{BS} = 11 km / h$
$t = 8.2 min$
b. ${\vec{v}}_{BS} = −5 km / h$
$t = 18 min$
c. ${\vec{v}}_{BS} =$ ${\vec{v}}_{BW} +$ ${\vec{v}}_{WS}$ $θ = 22 °$ west of north

Vectors V sub B W, V sub W S and V sub B S form a right triangle, with V sub B W as the hypotenuse. V sub B S points up. V sub W S points to the right. V sub B W points up and left, at an angle of theta to the vertical. V sub B S is the vector sum of v sub B W and V sub W S.

d. $| {\vec{v}}_{BS} | =$ $7.4 km / h$ $t = 6.5 min$
e. ${\vec{v}}_{BS} =$ $8.54 km / h,$ but only the component of the velocity straight across the river is used to get the time

Vectors V sub B W, V sub W S and V sub B S form a right triangle, with V sub B S as the hypotenuse. V sub B W points up. V sub W S points to the right. V sub B S points up and right, at an angle of theta to the vertical. V sub B S is the vector sum of v sub B W and V sub W S.

$t = 6.0 min$
Downstream = 0.3 km

4-77 §4.5 source ↗

A cyclist traveling southeast along a road at 15 km/h feels a wind blowing from the southwest at 25 km/h. To a stationary observer, what are the speed and direction of the wind?

Answer

${\vec{v}}_{A G} =$ ${\vec{v}}_{A C} +$ ${\vec{v}}_{C G}$
$| {\vec{v}}_{A C} | =$ $25 km / h | {\vec{v}}_{C G} | =$ $15 km / h | {\vec{v}}_{A G} | =$ $29.15 km / h$ ${\vec{v}}_{A G} =$ ${\vec{v}}_{A C} +$ ${\vec{v}}_{C G}$
The angle between ${\vec{v}}_{A C}$ and ${\vec{v}}_{A G}$ is $31 °,$ so the direction of the wind is $14 °$ north of east.

Vectors V sub A C, V sub C G and V sub A G form a triangle. V sub A C and V sub C G are at right angles. V sub A G is the vector sum of v sub A C and V sub C G.

4-79 §4.AP source ↗

A Formula One race car is traveling at 89.0 m/s along a straight track enters a turn on the race track with radius of curvature of 200.0 m. What centripetal acceleration must the car have to stay on the track?

Answer

a_{C} = 39.6 m / s^{2}

4-81 §4.AP source ↗

The driver of a car moving at 90.0 km/h presses down on the brake as the car enters a circular curve of radius 150.0 m. If the speed of the car is decreasing at a rate of 9.0 km/h each second, what is the magnitude of the acceleration of the car at the instant its speed is 60.0 km/h?

Answer

90.0 km / h =

25.0 m / s, 9.0 km / h =

2.5 m / s,

60.0 km / h =

16.7 m / s

a_{T} =

−2.5 {m/s}^{2}, a_{C} =

1.86 m / s^{2},

a =

3.1 m / s^{2}

4-83 §4.AP source ↗

An elephant is located on Earth’s surface at a latitude $λ .$ Calculate the centripetal acceleration of the elephant resulting from the rotation of Earth around its polar axis. Express your answer in terms of $λ,$ the radius $R_{E}$ of Earth, and time T for one rotation of Earth. Compare your answer with g for $λ = 40 ° .$

The earth is illustrated rotating about the vertical north south axis. The equator is shown as a horizontal circle at the earth’s surface, centered on the earth’s center. A second circle at the earth’s surface, parallel to the equator but north of it, is shown. This circle is at latitude lambda, meaning that the angle between the radius to this circle and to the equator is lambda.

Answer

The radius of the circle of revolution at latitude

λ

R_{E} cos λ .

The velocity of the body is

\frac{2 π r}{T} . a_{C} =

\frac{4 π^{2} R_{E} cos λ}{T^{2}}

for

λ =

40 °,

a_{C} =

0.26 % g

4-85 §4.AP source ↗

A propeller blade at rest starts to rotate from t = 0 s to t = 5.0 s with a tangential acceleration of the tip of the blade at

3.00 m / s^{2} .

The tip of the blade is 1.5 m from the axis of rotation. At t = 5.0 s, what is the total acceleration of the tip of the blade?

Answer

a_{T} = 3.00 m / s^{2}

v (5 s) =

15.00 m / s a_{C} =

150.00 m / s^{2} θ =

88.8 °

with respect to the tangent to the circle of revolution directed inward.

| \vec{a} | =

150.03 m / s^{2}

4-87 §4.AP source ↗

A particle’s centripetal acceleration is

a_{C} = 4.0 m / s^{2}

at t = 0 s where it is on the x-axis and moving counterclockwise in the xy plane. It is executing uniform circular motion about an axis at a distance of 5.0 m. What is its velocity at t = 10 s?

Answer

\vec{a} (t) =

− A ω^{2} cos ω t \hat{i} -

A ω^{2} sin ω t \hat{j}

a_{C} =

5.0 m ω^{2} ω =

0.89 rad / s

\vec{v} (t) =

−2.24 m / s \hat{i} -

3.87 m / s \hat{j}

4-89 §4.AP source ↗

A particle located initially at

(1.5 \hat{j} + 4.0 \hat{k}) m

undergoes a displacement of

(2.5 \hat{i} + 3.2 \hat{j} - 1.2 \hat{k}) m .

What is the final position of the particle?

Answer

{\vec{r}}_{1} =

1.5 \hat{j} +

4.0 \hat{k} {\vec{r}}_{2} =

Δ \vec{r} +

{\vec{r}}_{1} =

2.5 \hat{i} +

4.7 \hat{j} +

2.8 \hat{k}

4-91 §4.AP source ↗

A spaceship is traveling at a constant velocity of

\vec{v} (t) =

250.0 \hat{i} m / s

when its rockets fire, giving it an acceleration of

\vec{a} (t) =

(3.0 \hat{i} + 4.0 \hat{k}) m / s^{2} .

What is its velocity

5

s after the rockets fire?

Answer

v_{x} (t) =

265.0 m / s

v_{y} (t) =

20.0 m / s

\vec{v} (5.0 s) =

(265.0 \hat{i} + 20.0 \hat{j}) m / s

4-93 §4.AP source ↗

A long jumper can jump a distance of 8.0 m when he takes off at an angle of

45 °

with respect to the horizontal. Assuming he can jump with the same initial speed at all angles, how much distance does he lose by taking off at

30 ° ?

Answer

R = 1.07 m

4-95 §4.AP source ↗

A mountain biker encounters a jump on a race course that sends him into the air at

60 °

to the horizontal. If he lands at a horizontal distance of 45.0 m and 20 m below his launch point, what is his initial speed?

Answer

v_{0} = 20.1 m / s

4-97 §4.AP source ↗

A geosynchronous satellite orbits Earth at a distance of 42,250.0 km and has a period of 1 day. What is the centripetal acceleration of the satellite?

Answer

v = 3072.5 m / s

a_{C} =

0.223 m / s^{2}

4-99 §4.CP source ↗

World’s Longest Par 3. The tee of the world’s longest par 3 sits atop South Africa’s Hanglip Mountain at 400.0 m above the green and can only be reached by helicopter. The horizontal distance to the green is 359.0 m. Neglect air resistance and answer the following questions. (a) If a golfer launches a shot that is

40 °

with respect to the horizontal, what initial velocity must she give the ball? (b) What is the time to reach the green?

Answer

− 400.0 m =

v_{0 y} t -

4.9 t^{2} 359.0 m =

v_{0 x} t t =

\frac{359.0}{v_{0 x}} -

400.0 =

359.0 \frac{v_{0 y}}{v_{0 x}} -

4.9 {(\frac{359.0}{v_{0 x}})}^{2}

−400.0 =

359.0 tan 40 -

\frac{631,516.9}{v_{0 x}^{2}} \Rightarrow v_{0 x}^{2} =

900.6 v_{0 x} =

30.0 m / s v_{0 y} =

v_{0 x} tan 40 =

25.2 m / s

v = 39.2 m / s

, b.

t = 12.0 s

4-101 §4.CP source ↗

A truck is traveling east at 80 km/h. At an intersection 32 km ahead, a car is traveling north at 50 km/h. (a) How long after this moment will the vehicles be closest to each other? (b) How far apart will they be at that point?

Answer

{\vec{r}}_{T C} =

(−32 + 80 t) \hat{i} +

50 t \hat{j},

{| {\vec{r}}_{T C} |}^{2} =

{(−32 + 80 t)}^{2} +

{(50 t)}^{2}

2 r \frac{d r}{d t} =

2 (−32 + 80 t) (80) +

5000 t \frac{d r}{d t} =

\frac{160 (−32 + 80 t) + 5000 t}{2 r} =

0

17800 t =

5184 \Rightarrow t =

0.29 hr

,
b.

| {\vec{r}}_{T C} | = 17 km

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Problems and answers are adapted from University Physics Volume 1 by Samuel J. Ling, Jeff Sanny and William Moebs, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/university-physics-volume-1/pages/1-introduction. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.