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# 1.2 Position and Displacement

One general way of locating a particle (or particle-like object) is with a position vector , which is a vector that extends from a reference point (usually the origin of a coordinate system) to the particle. In the unit-vector notation, can be written

, (4-1)

where , , and are the vector components of , and the coefficients , , and are its scalar components.

The coefficients , , and give the particle's location along the coordinate axes and relative to the origin; that is, the particle has the rectangular coordinates ( , , ). For instance, Fig. 4-1 shows a particle with position vector

(4-2)

 Fig. 4-1
and rectangular coordinates . Along the axis the particle is 3 m from the origin, in the direction. Along the axis it is 2 m from the origin, in the direction. Along the axis it is 5 m from the origin, in the direction.

As a particle moves, its position vector changes in such a way that the vector always extends to the particle from the reference point (the origin). If the position vector changes, say, from to during a certain time interval, then the particle's displacement during that time interval is

(4-2)

Using the unit-vector notation, we can rewrite this displacement as

or as

(4-3)

where coordinates ( , , ) correspond to position vector and coordinates ( , , ) correspond to position vector . We can also rewrite the displacement by substituting for , for , and for :

(4-3)

Example 4-1

In Fig. 4-2, the position vector for a particle is initially

 Fig. 4-2
and then later is

.

What is the particle's displacement from to ?

Solution. The Key Idea is that the displacement is obtained by subtracting the initial position vector from the later position vector . That is most easily done by components:

This displacement vector is parallel to the plane, because it lacks any component, a fact that is easier to see in the numerical result than in Fig. 4-2.

Sample Problem 4-2

A rabbit runs across a parking lot on which a set of coordinate axes has been drawn. The coordinates of the rabbit's position as functions of time are given by

(4-5)

And , (4-6)

with in seconds and and in meters.

(a) At s, what is the rabbit's position vector in unit-vector notation and as a magnitude and an angle?

Solution. The Key Idea here is that the and coordinates of the rabbit's position, as given, are the scalar components of the rabbit's position vector . Thus, we can write

(We write rather than because the components are functions of , and thus is also.)

At s, the scalar components are

m,

m.

Thus, at

To get the magnitude and angle of , we can use a vector-capable calculator, or we can write

(Although has the same tangent as - 41°, study of the signs of the components of rules out 139°.)

(b) Graph the rabbit's path for to s.

Solution: We can repeat part (a) for several values of and then plot the results. Figure 4-36 shows the plots for five values of and the path connecting them. We can also use a graphing calculator to make a. parametric graph; that is, we would have the calculator plot у versus x, where these coordinates are given by Eqs. 4-5 and 4-6 as functions of time .

4-3 Average Velocity and Instantaneous Velocity

If a particle moves through a displacement in a time interval , then its average velocity is

Or (4-8)

This tells us the direction of must be the same as that of the displacement . Using Eq. 4-4, we can write Eq. 4-8 in vector components as

(4-9)

For example, if the particle in Sample Problem 4-1 moves from its initial position to its later position in 2.0 s, then its average velocity during that move is

When we speak of the velocity of a particle, we usually mean the particle's instantaneous velocity at some instant. This is the value that approaches in the limit as we shrink the time interval to 0 about that instant. Using the language of calculus, we may write as the derivative

(4-10)

Figure 4-4 shows the path of a particle that is restricted to the plane. As the particle travels to the right along the curve, its position vector sweeps to the right. During time interval , the position vector changes from to , and the particle's displacement is .

 Fig. 4-4
To find the instantaneous velocity of the particle at, say, instant (when the particle is at position 1), we shrink interval to 0 about . Three things happen as we do so: (1) Position vector in Fig. 4-4 moves toward so that shrinks toward zero. (2) The direction of (thus of ) approaches the direction of the tangent line to the particle's path at position 1. (3) The average velocity approaches the instantaneous velocity at .

In the limit as , we have and, most important here, takes on the direction of the tangent line. Thus, has that direction as well:

The direction of the instantaneous velocity of a particle is always tangent to the particle's path at the particle's position.

The result is the same in three dimensions: is always tangent to the particle's path.

To write Eq. 4-10 in unit-vector form, we substitute for from Eq. 4-1:

.

This equation can be simplified somewhat by writing it as

(4-11)

where the scalar components of are

(4-12)

For example, is the scalar component of along the axis. Thus, we can find the scalar components of by differentiating the scalar components of .

Figure 4-5 shows a velocity vector and its scalar x and у components. Note that is tangent to the particle's path at the particle's position. Caution: When a position vector is drawn, it is an arrow that extends from one point (a "here") to another point (a "there"). However, when a velocity vector is drawn as in Fig. 4-5, it does not extend from one point to another. Rather, it shows the instantaneous direction of travel of a particle located at the tail, and its length (representing the velocity magnitude) can be drawn to any scale.

Sample Problem 4-3

For the rabbit in Sample Problem 4-2, find the velocity at time s, in unit-vector notation and as a magnitude and an angle.

Solution There are two Key Ideas here: (1) We can find the rabbit's velocity by first finding the velocity components. (2) We can find those components by taking derivatives of the components of the rabbit's position vector. Applying the first of Eqs. 4-12 to Eq. 4-5, we find the x component of to be

(4-13)

At s, this gives m/s.

Similarly, applying the second of Eqs. 4-12 to Eq. 4-6, we find that the component is

At s, this gives m/s.

Equation 4-11 then yields

which is shown in Fig. 4-6, tangent to the rabbit's path and in the direction the rabbit is running at s.

To get the magnitude and angle of , either we use a vector-capable calculator or we follow Eq. 3-6 to write

(Although 50° has the same tangent as -130°, inspection of the signs of the velocity components indicates that the desired angle is in the third quadrant, given by 50° - 180° = -130°.)

4-4 Average Acceleration and Instantaneous Acceleration

When a particle's velocity changes from to in a time interval , its average acceleration during is

Or

If we shrink to zero about some instant, then in the limit approaches the instantaneous acceleration (or acceleration) at that instant; that is,

(4-16)

If the velocity changes in either magnitude or direction (or both), the particle must have an acceleration.

We can write Eq. 4-16 in unit-vector form by substituting for v*from Eq. 4-11 to obtain

We can rewrite this as

(4-17)

where the scalar components of are

(4-18)

We can rewrite this as

(4-17)

where the scalar components of are

, (4-18)

Sample Problem 4-4

For the rabbit in Sample Problems 4-2 and 4-3, find the acceleration at time s, in unit-vector notation and as a magnitude and an angle.

Solution: There are two Key Ideas here: (1) We can find the rabbit's acceleration by first finding the acceleration components. (2) We can find those components by taking derivatives of the rabbit's velocity components. Applying the first of Eqs. 4-18 to Eq. 4-13, we find the component of to be

m/s2

Similarly, applying the second of Eqs. 4-18 to Eq. 4-14 yields the component as

m/s2

We see that the acceleration does not vary with time (it is a constant) because the time variable does not appear in the expression for either acceleration component. Equation 4-17 then yields

which is shown superimposed on the rabbit's path in Fig. 4-8.

To get the magnitude and angle of , either we use a vector-capable calculator or we follow Eq. 3-6. For the magnitude we have

For the angle we have

For each description, determine whether the x and у components of the puck's acceleration are constant, and whether the acceler­ation a* is constant.

However, this last result, which is displayed on a calculator, indicates that is directed to the right and downward in Fig. 4-8. Yet, we know from the components above that must be directed to the left and upward. To find the other angle that has the same tangent as - 35°, but which is not displayed on a calculator, we add 180°:

-35° + 180° = 145c

This is consistent with the components of . Note that has the same magnitude and direction throughout the rabbit's run because, as we noted previously, the acceleration is constant.

Exercises

1. The position of an object moving only an -axis is given by , where is in meters and in seconds. (a) What is the position of the object at s? (b) What is the displacement’s between and s? (c) What is the average velocity for the time interval from and ?

2. (a) If particle’s position is given by , (where is in meters and in seconds), what its velocity at ? (b) Is it moving in positive or negative direction of just then? (c) What is its speed just then? (d) Is the speed larger or smaller at later time? (e) Is there ever an instant when the velocity is zero? (f) Is there a time after when the particle is moving in the negative direction of ?

3. The position of a particle moving along the axis is given in centimeters by , where is in seconds. Calculate (a) the average velocity during the time interval s to s; (b) the instantaneous velocity at s; (c) the instantaneous velocity at s; (d) the instantaneous velocity at s; and (e) the instantaneous velocity when the particle is midway between its positions at f s and s. (f) Graph versus and indicate your answers graphically.

4. A proton moves along the axis according to the equation , where is in meters and is in seconds. Calculate (a) average velocity of the proton during the first 3.0 s of its motion. (b) the instantaneous velocity of the proton at s, (c) the instantaneous acceleration of the proton at s.

5. An electron moving along the axis has a position given by m. where is in seconds. How far is the electron from the origin when it momentarily stops?

6. The position of a particle moving along the axis depends on the time according to the equation , where is in meters and in seconds, (a) What units must с and b have? Let there numerical values be 3.0 and 2.0, respectively. (b) At what time does the particle reach its maximum positive position? From and , (c) what distance does the particle move and (d) what is its displacement? At s what are (e) its velocity and (f) its acceleration?

Exercises

The and coordinates of a particle at any time are given by and , where and are in meters and in seconds. The acceleration of the particle at s is

(A) zero (B) 8 m/ s2 (C) 20 m/s2 (D) 40 m·s2

2-34 A jet-propelled motorcycle starts from rest, moves in a straight line with constant acceleration, and covers a distance of 64 m in 4 s.

1. What is the final velocity?

2. How much time was required to cover half the total distance?

3. What is the distance covered in one-half the total time?

4. What is the velocity when half the total distance has been covered?

5. What is the velocity after one-half the total time?

6. When will the instantaneous velocity equal the average velocitv for the 0-to-4 s time interval?

Q. 2.09. If the displacement of a body is proportional to square of time, state whether the body is moving with uniform velocity or uniform acceleration.

Q. 2.10. If the distance traveled by a body in time is given by , then what will be the acceleration of the body?

Q. 2.11. An object is covering distance in direct proportion to , where is the time elapsed.

(a) What conclusions might you draw about acceleration? Is it constant? Increasing? Decreasing?

(b) What might you conclude about the force acting on the object?

2. The position of a particle is given by , where t is in seconds and the coefficients have the proper units for to be in meters.

(a) Find and of the particle.

(b) Find the magnitude and direction of at s.

[Ans. (a) ; ; (b) 12.37 m/s; 76°]

Section 2—5 Velocity and Coordinate by Integration

2-20 The motion of a particle along a straight line is described by the function

x = (6 m) + (5 m-s-2)/2 - (1 m-s-4)/4.

Assume that t is positive.

1. Find the position, velocity, and acceleration at time t = 2 s.

1. During what time interval is the velocity positive?

2. During what time interval is x positive?

1. What is the maximum positive velocity attained by the particle?

2—21 The acceleration of a motorcycle is given by a - (1.2 m-s-3)/ - (0.12 m-s-4)/2. It is at rest at the origin

at time t 0.

Find its position and velocity as functions of time.

1. Calculate the maximum velocity it attains.

2-22 The acceleration of a bus is given by a = (2 m-s-3)t.

) If the bus's velocity at time t = 1 s is 5 m-s-1, what is its velocity at time ( = 2s?

b) If the bus's position at time t = 1 s is 6 m, what is its position at time t = 2 s?

56 The acceleration of an object suspended from a spring and oscillating vertically is , where is a constant and is the coordinate measured from the equilibrium position. Suppose that an object moving in this way is given an initial velocity at the coordinate . Find the expression for the velocity of the object as a function of its coordinate .

2-57 The motion of an object falling from rest in a resisting medium is described by the equation

where A and В are constants. In terms of A and B, find

1. the initial acceleration;

2. the velocity at which the acceleration becomes zero (the terminal velocity).

3. Show that the velocity at any given t is given by

2-58 After the engine of a moving motorboat is cut off, the boat has an acceleration in the opposite direction to its velocity and directly proportional to the square of its velocity. That is, , where is constant.

a) Show that the magnitude of the velocity at a time after the engine is cut off is given by

.

b) Show that the distance x traveled in a time is

c) Show that the velocity after traveling a distance is

As a numerical example, suppose the engine is cut off when the velocity is 6 m/s, and that the velocity decreases to 3 m/s in a time of 15 s.

d) Find the numerical value of the constant , and the unit in which it is expressed.

e) Find the acceleration at the instant the engine is cut off.

f) Calculate x, v, and a at 10-s intervals for the first 60 s after the engine is cut off. Sketch graphs of x, v. and a as

functions oi t, tor the first 60 s of the motion.

4-7 Uniform Circular Motion

A particle is in uniform circular motion if it travels around a circle or a circular arc at constant (uniform) speed. Although the speed does not vary, the particle is accelerating. That fact may be surprising because we often think of acceleration (a change in velocity) as an increase or decrease in speed. However, actually velocity is a vector, not a scalar. Thus, even if a velocity changes only in direction, there is still an acceleration, and that is what happens in uniform circular motion.

Figure 4-18 shows the relation between the velocity and acceleration vectors at various stages during uniform circular motion. Both vectors have constant magnitude as the motion progresses, but their directions change continuously. The velocity is always directed tangent to the circle in the direction of motion. The acceleration is always directed radially inward. Because of this, the acceleration associated with uniform circular motion is called a centripetal (meaning "center seeking") acceleration. As we prove next, the magnitude of this acceleration is

(centripetal acceleration) (4-32)

where is the radius of the circle and is the speed of the particle.

In addition, during this acceleration at constant speed, the particle travels the circumference of the circle (a distance of ) in time

(period). (4-33)

is called the period of revolution, or simply the period, of the motion. It is, in general, the time for a particle to go around a closed path exactly once.

Proof of Eq. 4-32

To find the magnitude and direction of the acceleration for uniform circular motion, we consider Fig. 4-19. In Fig. 4-l9a, particle p moves at constant speed around a circle of radius . At the instant shown, p has coordinates and .

Recall from Section 4-3 that the velocity of a moving particle is always tangent to the particle's path at the particle's position. In Fig. 4-19a, that means is perpendicular to a radius drawn to the particle's position. Then the angle that makes with a vertical at p equals the angle that radius makes with the axis.

The scalar components of are shown in Fig. 4-19b. With them, we can write the velocity as

(4-34)

Now, using the right triangle in Fig. 4-19a, we can replace with and with to write

(4-35)

To find the acceleration of particle p, we must take the time derivative of this equation. Noting that speed and radius do not change with time, we obtain

(4-36)

Now note that the rate at which changes is equal to the velocity component . Similarly, , and, again from Fig. 4-19b, we see that and . Making these substitutions in Eq. 4-36, we find

(4-37)

This vector and its components are shown in Fig. 4-19c. Following Eq. 3-6, we find that the magnitude of is

as we wanted to prove. To orient , we can find the angle shown in Fig. 4-19c:

Thus, , which means that is directed along the radius of Fig. 4-19a toward the circle's center, as we wanted to prove.

Sample Problem 4-9

"Top gun" pilots have long worried about taking a turn too tightly. As a pilot's body undergoes centripetal acceleration, with the head toward the center of curvature, the blood pressure in the brain decreases, leading to loss of brain function.

There are several warning signs to signal a pilot to ease up: when the centripetal acceleration is or , the pilot feels heavy. At about , the pilot's vision switches to black and white and narrows to "tunnel vision." If that acceleration is sustained or increased, vision ceases and, soon after, the pilot is unconscious - a condition known as -LOC for " -induced loss of consciousness." What is the centripetal acceleration, in units, of a pilot flying an F-22 at speed = 2500 km/h (694 m/s) through a circular arc with radius of curvature = 5.80 km?

SOLUTION: The Key Idea here is that although the pilot's speed is constant, the circular path requires a (centripetal) acceleration, with magnitude given by Eq. 4-32: