Two-dimensional taken from . A particle moves on a horizontal surface according to in meters and seconds. When, if ever, is the x component of the velocity zero? The y component of the velocity zero? The magnitude of the velocity zero?

Calculations: Taking the time derivative of each component of , we find with in meters per second. The x component is 3 m/s and is never zero. The y component is and is zero when . The magnitude of the velocity is which can never be zero because the x component is always 3 and the square of the y component cannot be a negative number.

Variations:

1.   When is the x component of the acceleration zero? Answer: always 2.   When is the y component of the acceleration zero? Answer: never 3.   When is the magnitude of the acceleration zero? Answer: never

Two-dimensional taken from . A particle with velocity at undergoes acceleration while moving in an xy plane. What is the particle’s velocity at , in unit-vector notation and in magnitude-angle notation?

Calculations: Writing Eq. 2-22 for this situation, we have

	(Answer)

The magnitude of is

	(Answer)

The direction of is determined by the angle between and the positive direction of the x axis. We find with

	(Answer)

Drawing and its components confirms that the angle is , not .

Variations:

1.   When, if ever, is the magnitude of the acceleration zero? Answer: never 2.   When, if ever, is the x velocity component zero? Answer: 1 s and −1 s 3.   When, if ever, is the magnitude of the velocity zero? Answer: never

Collision in two-dimensional motion. Figure 4-C shows two particles at time . Particle A is passing through the origin with velocity , and particle B is passing through coordinates (12 m, −8.0 m) with constant velocity . What constant acceleration of A is required such that A travels along the x axis to collide with B?

FIGURE 4-C

Calculations: Because A is to travel along the x axis, the collision point must be on the x axis, at coordinates (12 m, 0).To reach that point, particle B travels parallel to the y axis at constant velocity. Of the constant-acceleration equations, we choose Eq. 2-15, rewriting it for motion parallel to the y axis: Substituting , , , and , and then solving for the time for B to reach the collision point, we find that .

For A we use the same constant-acceleration equation except we now write it for motion parallel to the x axis: Substituting , , , and , we find that A must have an acceleration along the x axis of . Thus,

	(Answer)

Variations:

1.   If the data are as given in the problem statement but A does not undergo an acceleration, does it arrive at the point (12 m, 0) before or after B? Answer: before 2.   What velocity parallel to the x axis must A have at if it is to collide with B without having to undergo an acceleration? Answer:

Projectile, horizontal launch. In Fig. 4-D, a movie stuntman is to run across and directly off a rooftop, to land on the roof of the next building. Can he land there if he runs at 4.5 m/s?

FIGURE 4-D

Calculations: The initial velocity is horizontal () and has magnitude . For him to land on the second roof, his horizontal displacement during the flight must be at least 6.2 m. We should be able to find the actual value of for the given initial speed by using Eq. 4-21,

	(01)

but for that we need his time of flight t.

To find t, we consider the vertical motion and use Eq. 4-22:

	(02)

Here his vertical displacement during the flight is −4.8 m. Putting this and other known values into Eq. 02 gives us which yields . Using this in Eq. 01, we have Thus, with his given running speed, he cannot land on the roof of the next building.

Determining projectile launch using mid-flight data. A particle is launched into projectile motion over level ground. At point A, 3.0 m above ground level, the particle has velocity . What was the angle between its path and the positive direction of the x axis at the launch point?

Calculations: Because the horizontal velocity component does not change, its value at the launch point must be the same as at point A, namely . Thus, .

Because the vertical acceleration is constant, we can use the constant-acceleration equations of Chapter 2. We use Eq. 2-16 written for y motion in the form of Eq. 4-24:

Substituting , , and , we solve for the quantity and use the positive root when taking the square root. That quantity is the vertical velocity at launch. Thus, . At launch, the velocity and the path have the angle

	(Answer)

Variations:

1.   What is the launch speed? Answer: 10.9 m/s 2.   The particle is at a height of 3.0 m at one other instant during its flight. What is its velocity at that other instant? Answer:

Projectile on a slope. In Fig. 4-F, a projectile is launched at speed and at angle up over sloped ground that rises 1.00 m for every horizontal 3.00 m. At what xy coordinates does the projectile land?

FIGURE 4-F

Calculations: To determine where the projectile hits the sloped ground, we first write equations y(x) for the height as a function of horizontal position, for both the projectile’s path and the ground. Then we solve the equations simultaneously for x. For the projectile, we use Eq. 4-25:

	(01)

Substituting and , this becomes

	(02)

For the hill, we write a linear equation in the form , where the slope is and the y intercept is :

	(03)

Solving Eqs. 02 and 03 simultaneously by eliminating y and then solving the resulting quadratic equation for x, we find

	(Answer)

Substituting this result into either Eq. 02 or 03 gives us

	(Answer)

Variation: If the slope of the ground is decreased, does the x coordinate of the landing point increase, decrease, or remain the same?

Answer: increase

Uniform circular motion. A satellite is in circular Earth orbit, at altitude above Earth’s surface. There the free-fall acceleration g is . What is the orbital speed v of the satellite?

Calculations: We can find v from Eq. 4-34 , with and with , where is Earth’s radius . With those substitutions, we have Solving for v gives

	(Answer)

You can show that this is equivalent to 28 000 km/h and that (from Eq. 4-35) the satellite would take 1.47 h to complete one orbital revolution; that is, the period T of the motion is 1.47 h.

Uniform circular motion. A particle undergoes uniform circular motion on a horizontal xy plane. At time , it moves through coordinates (3.0 m, 0) with velocity . At , it moves through (11.0 m, 0) with velocity . In unit vector notation, what is its acceleration at ?

Calculations: Because the two given velocity vectors are in opposite directions, they must be on opposite sides of the circular path. Because the vectors are tangent to the circle and directed parallel to the y axis, they must be on the left side and right side of the circle. Thus at , the particle is on the left side at coordinates (3.0 m, 0), as indicated in Fig. 4-H. At , it is on the right side at coordinates (11.0 m, 0). The center of the circle must be midway between these points, at coordinates (7.0 m, 0), which means the radius is .

FIGURE 4-H

The speed is the magnitude of , which is 6.0 m/s. We then have At , the particle is at the top of the circle, midway between the left side and the right side, and thus it must be at coordinates (7.0 m, 4.0 m). Because the acceleration vector is always directed toward the center of the circle, at it must be directed down toward (7.0 m, 0). Therefore,

	(Answer)

Variations:

1.   What is when ? Answer: 3.3 m/s 2.   What is a when ? Answer:

One-dimensional relative motion. A boat moves downstream in the positive direction of an x axis. The boat’s velocity relative to the shore is , and the water’s velocity relative to the shore is . A child walks upstream from the front to the rear of the boat with a velocity of relative to the boat. What is the child’s velocity relative to the water?

Calculations: The velocity of the boat relative to the shore is , and the velocity of the water relative to the shore is . Thus, the velocity of the boat relative to the water is where we use the fact that is the reverse of . Notice how we expand the initial subscript bw to become the subscripts bs and sw.

The velocity of the child relative to the boat is . We can now write that the velocity of the child relative to the water is

	(Answer)

Variations:

1.   What is the velocity of the shore relative to the child? Answer: 2.   If the child accelerates at relative to the boat, what is the child’s acceleration relative to the shore? Answer:

Two-dimensional relative motion. In Fig. 4-Ja, a bat detects an insect (lunch) while the two are flying. The bat has velocity relative to the ground, and the insect has velocity relative to the ground. What is the velocity of the insect relative to the bat, in unit-vector notation and as a magnitude and an angle?

FIGURE 4-J

Calculations: The moving particle P is now the insect (I), frame A is now the ground (G), and frame B is now the bat (B). We first construct a sentence that relates the three vectors: Now we can draw this relation as in Fig. 4-Jb and write it as or We can evaluate the right side of this last equation directly on a vector-capable calculator by keying in the given vectors. Alternatively, we can resolve the given vectors into vector components and then add and . We have and Thus,

	(Answer)

Using the components in the Pythagorean theorem and an inverse tangent, we find that has and

	(Answer)

Average velocity using unit vectors. At time , a particle is located by position vector . At , it is located by . What is the particle’s average velocity in unit-vector notation and in magnitude-angle notation?

Calculations: The displacement is Dividing this by the elapsed time of 2.0 s gives us the average velocity during that time interval:

	(Answer)

To find the magnitude of the average velocity, we insert the components in the Pythagorean theorem for three dimensions, where the third component is entered as the third squared term under the square root:

	(Answer)

To find the orientation of the average velocity, first note that the vector lies in the xz plane (because it does not have a y component). We can draw the components in a head-to-tail arrangement as shown in Fig. 4-K. The angle relative to the positive direction of the x axis is then given by :

	(Answer)

measured counterclockwise in Fig. 4-K.

FIGURE 4-K

Average acceleration using unit vectors. At time , a particle has velocity . At , it has velocity . What is the particle’s average acceleration in unit-vector notation and in magnitude-angle notation?

Calculations: The change in velocity is (being very careful about the unit vectors) Dividing this by the elapsed time of 2.0 s gives us the average velocity during that time interval:

	(Answer)

To find the magnitude of the average acceleration, we insert the components in the Pythagorean theorem for three dimensions, where the third component is entered as the third squared term under the square root:

	(Answer)

To find the orientation of the average velocity, first note that the vector lies in the xz plane (because it does not have a y component). We can draw the components in a head-to-tail arrangement as shown in Fig. 4-L. The angle relative to the positive direction of the x axis is then given by :

	(Answer)

measured counterclockwise in Fig. 4-L.

FIGURE 4-L