Unit 1. Kinematics 1.1 Scalars and Vectors All physical quantities can be categorized as either a vector quantity or a scalar quantity. A

vector has both direction and magnitude (size). A scalar can be completely specified by its

magnitude with appropriate units; it has no direction.

Displacement, velocity, acceleration, force and momentum are vector quantities. Distance, time,

speed, area, volume, mass, temperature, energy, power, pressure and charge are scalar

quantities. If the temperature of an object is 25°C, that information completely specifies the

temperature of the object; no direction is required. Scalar quantities can be manipulated with the

rules of algebra.

An example of a vector quantity is displacement. Suppose a particle moves from some point A

to some point B along a straight path as shown in Figure 3.4. We represent this displacement by

drawing an arrow from A to B, with the tip of the arrow pointing away from the starting point.

Figure 1.1 As a particle moves from A to B along an arbitrary path represented by the broken line, its displacement is a vector quantity shown by the arrow drawn from A to B.

The direction of the arrowhead represents the direction of the displacement, and the length of

the arrow represents the magnitude of the displacement. If the particle travels along some other

path from A to B such as shown by the broken line in Figure 1.1, its displacement is still the

arrow drawn from A to B. Displacement depends only on the initial and final positions, so the

displacement vector is independent of the path taken by the particle between these two points.

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Tip

Tail

When a vector quantity is written, it is often represented with an arrow over the letter ( A⃗).

Another common notation for vectors with which you should be familiar is a simple boldface

character: A. The magnitude of the vector A⃗ is written as A. The magnitude of a vector has

physical units, such as meters for displacement or meters per second for velocity. The

magnitude of a vector is always a positive number.

Vectors can be represented graphically as arrows. The vector’s magnitude is equal to the length

of the arrow, and its direction corresponds to where the arrow is pointing. Physicists commonly

refer to the point of a vector as its tip and the base as its tail.

1.1.1 Some Properties of Vectors

(I) Equality of Two VectorsTwo vectors A⃗ and B⃗ may be defined to be equal if they have the same magnitude and if they

point in the same direction. For example, all the vectors in Figure 1.2 are equal even though they

have different starting points. This property allows us to move a vector to a position parallel to

itself in a diagram without affecting the vector.

Figure 1.2 These four vectors are equal because they have equal lengths and point in the same direction.(II) Adding Vectors(a) Tip-to-Tail Method

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The rules for adding vectors are conveniently described by a graphical method. To add vector B⃗

to vector A⃗ , first draw vector A⃗ on graph paper, with its magnitude represented by a convenient

length scale, and then draw vector B⃗ to the same scale, with its tail starting from the tip of A⃗ , as

shown in Figure 1.3. The resultant vector R⃗ = A⃗ + B⃗ is the vector drawn from the tail of A⃗ to the

tip of B⃗. This technique for adding vectors is often called the “head to tail method.”

Figure 1.3 Addition of two vectors using head to tail method.

When two vectors are added, the sum is independent of the order of the addition. This property,

which can be seen from the geometric construction in Figure 1.4, is known as the commutative

law of addition: A⃗ + B⃗ = B⃗ + A⃗.

Figure 1.4 This construction shows that A⃗ + B⃗ = B⃗ + A⃗ or, in other words, that vector addition is commutative.(b) Parallelogram Method

To add A⃗ and B⃗ using the parallelogram method, place the tail of B⃗ so that it meets the tail of A⃗.

Take these two vectors to be the first two adjacent sides of a parallelogram, and draw in the

remaining two sides. The vector sum, R⃗ = A⃗ + B⃗, extends from the tails of A⃗ and B⃗ across the

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diagonal to the opposite corner of the parallelogram. This is shown in Figure 1.5. If the vectors

are perpendicular and unequal in magnitude, the parallelogram will be a rectangle. If the

vectors are perpendicular and equal in magnitude, the parallelogram will be a square.

Figure 1.5 Addition of two vectors using head to tail method.

(III) Negative of a VectorThe negative of the vector A⃗ is defined as the vector that when added to A⃗ gives zero for the

vector sum. That is, A⃗ + −⃗A = 0. The vectors A⃗ and −⃗A have the same magnitude but point in

opposite directions. This is shown in Figure 1.6.

Figure 1.6 Negative of a vector.(IV) Subtracting VectorsThe operation of vector subtraction makes use of the definition of the negative of a vector. We

define the operation A⃗ - B⃗ as vector −⃗B added to vector A⃗

The geometric construction for subtracting two vectors in this way is illustrated in Figure 1.7.

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Figure 1.7 Subtracting two vectors.

Another way of looking at vector subtraction is to notice that the difference vector C⃗ = A⃗ - B⃗

between two vectors A⃗ and B⃗ is what you have to add to the second vector B⃗ to obtain the first

vector A⃗. In this case, as Figure 1.8 shows, the vector C⃗ = A⃗ - B⃗ points from the tip of the second

vector to the tip of the first.

Figure 1.8 A second way of looking at vector subtraction(V) Components of a Vector The projections of a vector along coordinate axes are called the components of this vector or its

rectangular components. Any vector can be completely described by its components.

Consider a vector A⃗ lying in the xy plane and making an arbitrary angle θ with the positive x

axis as shown in Figure 1.9 a.

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Figure 1.9 (a) A vector A⃗ lying in the xy plane can be represented by its component vectors A⃗ x and A⃗ y. (b) The y-component vector A⃗ y can be moved to the right so that it adds to A⃗ x.

The vector sum of the component vectors is A⃗ . These three vectors form a right triangle.

This vector can be expressed as the sum of two other component vectors A⃗x , which is parallel to

the x axis, and A⃗ y , which is parallel to the y axis. From Figure 3.12b, we see that the three

vectors form a right triangle and that A⃗ = A⃗ x + A⃗ y . We shall often refer to the “components of a

vector A⃗ ,” written Ax and Ay (without the boldface notation). The component Ax represents the

projection of A⃗ along the x-axis, and the component Ay represents the projection of A⃗ along the

y-axis. The components of a vectors can be considered as the effect of the vector along the x

and y directions respectively. There are two cases:

(a) If the magnitude of the vector (A) and the angle (θ) with x-axis are known, then Ax and Ay

can be calculated as follows:

From Figure 3.12 and the definition of sine and cosine, we see that cos θ = Ax/A and that sin θ

= Ay /A. Hence, the components of A⃗ are:

(1.1)(1.2)

(b) If the magnitudes of the components Ax and Ay are known, then the magnitude of the vector

(A) and the angle (θ) with x-axis can be calculated as follows:

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The magnitudes of Ax and Ay are the lengths of the two sides of a right triangle with a

hypotenuse of length A. Therefore, the magnitude and direction of A⃗ are related to its

components through the expressions:

(1.3)

(1.4)

Example 1.1:

1.2 Motion in One DimensionThe study of motion and of physical concepts such as force and mass is called dynamics. The

part of dynamics that describes motion without regard to its causes is called kinematics.

In this section the focus is on kinematics in one dimension: motion along a straight line. This

kind of motion involves the concepts of displacement, velocity, and acceleration. Here, we use

these concepts to study the motion of objects undergoing constant acceleration.

1.2.1 Position and DisplacementTo locate an object means to find its position relative to some reference point, often the origin

(or zero point) of an axis such as the x-axis in Figure 1.10. The positive direction of the axis is

in the direction of increasing numbers (coordinates), which is to the right in Figure 1.10.The

opposite is the negative direction.

For example, a particle might be located at x = 5 m, which means it is 5 m in the positive

direction from the origin. If it were at x= -5 m, it would be just as far from the origin but in the

opposite direction. A plus sign for a coordinate need not be shown, but a minus sign must

always be shown.

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Figure 1.10 Position is determined on an axis that is marked in units of length (here meters) and that extends indefinitely in opposite directions. The axis name, here x, is always on the positive side of the origin.

The change from an initial position xi to a final position xf is called the displacement x of an object and is given by:

(1.5)

(The symbol , the Greek uppercase delta, represents a change in a quantity, and it means the

final value of that quantity minus the initial value.)

The SI unit of displacement is meter (m).

The gecko in Figure 1.11 moves from left to right along the x-axis from an initial position,

xi = 24 cm , to a final position, xf = 85 cm. The gecko’s displacement is the difference between

its final and initial coordinates, or, x xf − xi = 85 cm − 24 cm = 61 cm.

Figure 1.11 A gecko moving along the x-axis from xi to xf undergoes a displacement of Δx = xf − xi.

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Now suppose the gecko runs up a tree, as shown in Figure 1.12. In this case, we place the

measuring stick parallel to the tree. The measuring stick can serve as the y-axis of our

coordinate system. The gecko’s initial and final positions are indicated by yi and yf,

respectively, and the gecko’s displacement is denoted as Δy.

y = yf - yi (1.6)

Figure 1.12 When the gecko is climbing a tree, the displacement is measured on the y-axis. Again, the gecko’s position is determined by the position of the same point on its body.

Displacement does not always tell you the distance an object

has moved. For example, what if the gecko in Figure 2.3 runs

up the tree from the 20 cm marker (its initial position) to the 80

cm marker. After that, it retreats down the tree to the 50 cm

marker (its final position). It has traveled a total distance of 90

cm. However, its displacement is only 30 cm (yf - yi = 50 cm -

20 cm = 30 cm). If the gecko were to return to its starting point

(20 cm mark), its displacement would be zero because its initial

position and final position would be the same.Displacement can be positive or negative.

Displacement also includes a description of the direction of motion. In one-dimensional motion,

there are only two directions in which an object can move, and these directions can be described

as positive or negative. Unless otherwise stated, the right (or east) will be considered the

positive direction and the left (or west) will be considered the negative direction. Similarly,

upward (or north) will be considered positive, and downward (or south) will be considered

negative. Figure 1.13 gives examples of determining displacements for a variety of situations.

Displacement is an example of a vector quantity, which is a quantity that has both a direction

and a magnitude.

(1) Its magnitude is the distance (such as the number of meters) between the original and final

positions.

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(2) Its direction, from an original position to a final position, can be represented by a plus sign

or a minus sign if the motion is along a single axis.

Figure 1.13 Displacements for a variety of situations.

Note: What is the difference between the displacement and the distance travelled?

The distance travelled between two points A and B equals the length of path travelled

between A and B. The magnitude of displacement is the length of the straight line joining A

and B. While displacement can be positive or negative, distance is always positive. Figure 1.14

shows an example of the difference between the distance travelled and the displacement.

Figure 1.14 Distance versus displacement

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Example 1.1:

Taking the Earth’s orbit to be a circle of radius 1.5× 108 km, determine the displacement

magnitude of the Earth and the distance it covers in a) half a year, b) one year.

Solution:

a) In half year: Displacement = Diameter = 2r = 3×108 kmDistance = ½ circle perimeter = ½ × 2πr = 4.7×108 kmb) In one year: Displacement = 0 km (initial position = final position)Distance = circle perimeter = 2πr = 9.4×108 km

Figure 1-15 Example 2.1

Example 1.2:A ball is thrown upward from the top of a building with an initial speed of 20.0 m/s, as in

Figure. The point of release is 45.0 m above the ground. What are the horizontal and vertical

displacements of the ball?

Figure 1-16 Displacement in two-dimensional motion.

Solution:xi = 0 m to xf = 73 m, then x = 73 – 0 = 73 m.

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yi = 45 m to yf = 0 m, then y = 0 – 45 = -45 m. (The negative sign indicated that the ball moves downward in the y-direction).

1.2.2 Average Velocity and Average Speed

Figure 1.17 The average velocity of this car tells you how fast and in which direction it is moving.

Consider the car in Figure 1.17. The car is moving along a highway in a straight line (the x-

axis). Suppose that the positions of the car are xi at time ti and xf at time tf. In the time interval

Δt = tf − ti , the displacement of the car is Δx = xf − xi . The average velocity, vavg, is defined as

the ratio of displacement divided by the time interval during which the displacement occurred.

In SI, the unit of velocity is meters per second (m/s).

(1.7)

The average velocity of an object can be positive or negative, depending on the sign of the

displacement. (The time interval is always positive.) As an example, consider a car trip to a

friend’s house 370 km to the west (the negative direction) along a straight highway. If you left

your house at 10 A.M. and arrived at your friend’s house at 3 P.M., your average velocity

would be as follows:

This value is an average. You probably did not travel exactly 74 km/h at every moment. You

may have stopped to buy gas or have lunch. At other times, you may have traveled more slowly

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as a result of heavy traffic. To make up for such delays, when you were traveling slower than 74

km/h, there must also have been other times when you traveled faster than 74 km/h.

Velocity is not the same as speed.In everyday language, the terms speed and velocity are used interchangeably. In physics,

however, there is an important distinction between these two terms. As we have seen, velocity

describes motion with both a direction and a numerical value (a magnitude) indicating how fast

something moves. However, speed has no direction, only magnitude. An object’s average speed

is equal to the distance traveled divided by the time interval for the motion.

(1.8)

1.2.3 Acceleration When a shuttle bus approaches a stop, the driver begins to apply the brakes to slow down 5.0 s

before actually reaching the stop. If the magnitude of the velocity changes from 10 m/s to 0 m/s

over a time interval of 5.0 s, then the change in velocity Δv = vf - vi = 0 m/s - 10 m/s =

-10 m/s. This change took 5 s time interval to take place, then Δt = tf - ti = 5 s. Then, the

change in velocity in one second time interval is Δv / Δt = (-10 m/s) / (5 s) = - 2 (m/s) ∕s

= -2 m/s2. This is the rate of change of velocity or the acceleration (a) of the bus. Then

a = -2 m/s2 means that the velocity decreases by a 2 m/s every second.

Sometimes, however, the shuttle stops much more quickly. For example, if the driver slams on

the brakes to avoid hitting a dog, the bus slows from 10 m/s to 0 m/s in just 2 s. Clearly, these

two stops are very different, even though the shuttle’s velocity changes by the same amount in

both cases. What is different in these two examples is the time interval during which the change

in velocity occurs. As you can imagine, this difference has a great effect on the motion of the

bus, as well as on the comfort and safety of the passengers. A sudden change in velocity feels

very different from a slow, gradual change. The quantity that describes the rate of change of

velocity in a given time interval is called acceleration. The magnitude of the average

acceleration is calculated by dividing the total change in an object’s velocity by the time

interval in which the change occurs.13

Consider a car moving along a straight highway as in Figure 1.18. At time t i it has a velocity of

vi, and at time tf its velocity is vf , with Δv = vf - vi and Δt = tf - ti, the average acceleration aavg is

(1.9)

Figure 1.18 A car moving to the right accelerates from a velocity of vi to a velocity of vf in the time interval Δt = tf - ti.

The units of acceleration in SI are meters per second per second, which is written as meters per second squared, as shown below. When measured in these units, acceleration describes how much the velocity changes in each second.

For example, suppose the car shown in Figure 1.18 accelerates from an initial velocity of vi = 12 m/s to a final velocity of vf = 20 m/s in a time interval of 2 s. The change in velocity Δv = 20 – 12 = 8 m/s. This change takes place during time interval Δt = 2 s. Then to find the change in velocity in 1 s (the acceleration) we divide Δv by Δt, then:aavg = 8 (m/s) / 2 (s) = 4 (m/s) / s or 4 m/s2. This means that on average, the magnitude of the car’s velocity increases by 4 m/s every second.

Acceleration is a vector quantity has both magnitude and. Its algebraic sign represents its direction on an axis just as for displacement and velocity; that is, acceleration with a positive value is in the positive direction of an axis, and acceleration with a negative value is in the negative direction.

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Example 1.3:

A shuttle bus slows down with an average acceleration of −1.8 m/s2. How long does it take the bus to slow from 9.0 m/s to a complete stop?

Solution

For the case of motion in a straight line, the direction of the velocity of an object and the

direction of its acceleration are related as follows: When the object’s velocity and

acceleration are in the same direction, the speed of the object increases with time. When

the object’s velocity and acceleration are in opposite directions, the speed of the object

decreases with time.

To clarify this point, suppose the velocity of a car changes from -10 m/s to -20 m/s in a time

interval of 2 s. The minus signs indicate that the velocities of the car are in the negative x-

direction; they do not mean that the car is slowing down! The average acceleration of the car in

this time interval is:

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The minus sign indicates that the acceleration vector is also in the negative x-direction. Because

the velocity and acceleration vectors are in the same direction, the speed of the car must

increase as the car moves to the left. Positive and negative accelerations specify directions

relative to chosen axes, not “speeding up” or “slowing down.” The terms speeding up or

slowing down refer to an increase and a decrease in speed, respectively.

Velocity can be interpreted graphically.

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The velocity of an object can be determined if the object’s position is known at specific times along its path. One way to determine this is to make a graph of the motion. Figure 2.10 represents such a graph.

FIGURE 2.10 Position-Time Graph The motion of an object moving with constant velocity will provide a straight-line graph of position versus time. The slope of this graph indicates the velocity.

Notice that time is plotted on the horizontal axis and position is plotted on the vertical axis. The object moves 4.0 m in the time interval between t = 0 s and t = 4.0 s. Likewise, the object moves an additional 4.0 m in the time interval between t = 4.0 s and t = 8.0 s. From these data, we see that the average velocity for each of these time intervals is +1.0 m/s (because v avg = Δx/Δt = 4.0 m/4.0 s). Because the average velocity does not change, the object is moving with a constant velocity of +1.0 m/s, and its motion is represented by a straight line on the position-time graph.For any position-time graph, we can also determine the average velocity by drawing a straight line between any two points on the graph. The slope of this line indicates the average velocity between the positions and times represented by these points.

A compact way to describe position is with a graph of position x plotted as a function of time t, a graph of x(t). (The notation x(t) represents a function x of t, not the product x times t.) As a simple example, Fig. 2-5 shows the position function x(t) for a stationary armadillo (which we treat as a particle) over a 7 s time interval. The animal’s position stays at x = -2 m.

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Figure 2-5 The graph of x(t) for an armadillo that is stationary at x = -2 m. The value of x is -2 m for all times t.

Figure 2-6 is more interesting, because it involves motion. The armadillo is apparently first noticed at t = 0 when it is at the position x = -5 m. It moves toward x = 0, passes through that point at t = 3 s, and then moves on to increasingly larger positive values of x. Figure 2-6 also depicts the straight-line motion of the armadillo (at three times) and is something like what you would see.

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Figure 2-6 The graph of x(t) for a moving armadillo. The path associated with the graph is also shown, at three times.

The graph in Fig. 2-6 is more abstract, but it reveals how fast the armadillo moves. Actually, several quantities are associated with the phrase “how fast.” One of them is the average velocity vavg, which is the ratio of the displacement Δx that occurs during a particular time interval Δt to that interval:

(2.3)

The notation means that the position is xi at time ti and then xf at time tf.A common unit for vavg

is the meter per second (m/s).

On a graph of x versus t, vavg is the slope of the straight line that connects two particular points on the x(t) curve: one is the point that corresponds to xf and tf, and the other is the point that corresponds to xi and ti. Like displacement, vavg has both magnitude and direction (it is another vector quantity). Its magnitude is the magnitude of the line’s slope. A positive vavg (and slope) tells us that the line slants upward to the right; a negative vavg (and slope) tells us that the line slants downward to the right.The average velocity vavg always has the same sign as the displacement Δx because Δt in Eq. 2-3 is always positive.

Figure 2-7 shows how to find vavg in Fig. 2-6 for the time interval t = 1 s to t = 4 s. We draw the straight line that connects the point on the position curve at the beginning of the interval and the point on the curve at the end of the interval. Then we find the slope Δx/Δt of the straight line. For the given time interval, the average velocity is:

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Figure 2-7 Calculation of the average velocity between t = 1 s and t = 4 s as the slope of the line that connects the points on the x(t) curve representing those times.

Figure 2.10 represents straight-line graphs of position versus time for three different objects.

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FIGURE 2.10 Position-Time Graphs These position-versus-time graphs show that Object 1 moves with a constant positive velocity. Object 2 is at rest. Object 3 moves with a constant negative velocity.FIGURE 1.7

Object 1 has a constant positive velocity because its position increases uniformly with time. Thus, the slope of this line is positive. Object 2 has zero velocity because its position does not change (the object is at rest). Hence, the slope of this line is zero. Object 3 has a constant negative velocity because its position decreases with time. As a result, the slope of this line is negative.

Instantaneous velocity may not be the same as average velocity.Now consider an object whose position-versus-time graph is not a straight line, but a curve, as in Figure 2.11. The object moves through larger and larger displacements as each second passes. Thus, its velocity increases with time.

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FIGURE 2.11 The instantaneous velocity at a given time can be determined by measuring the slope of the line that is tangent to that point on the position-versus-time graph.

For example, between t = 0 s and t = 2.0 s, the object moves 8.0 m, and its average velocity in this time interval is 4.0 m/s (because vavg = 8.0 m /2.0 s). However, between t = 0 s and t = 4.0 s, it moves 32 m, so its average velocity in this time interval is 8.0 m/s (because vavg = 32 m/4.0 s). We obtain different average velocities, depending on the time interval we choose. But how can we find the velocity at an instant of time?To determine the velocity at some instant, such as t = 3.0 s, we study a small time interval near that instant. As the intervals become smaller and smaller, the average velocity over that interval approaches the exact velocity at t = 3.0 s. This is called the instantaneous velocity. One way to

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determine the instantaneous velocity is to construct a straight line that is tangent to the position-versus-time graph at that instant. The slope of this tangent line is equal to the value of the instantaneous velocity at that point. For example, the instantaneous velocity of the object in Figure 2.11 at t = 3.0 s is 12 m/s.

_

Figure 2.9 gives plots of the position, velocity, and acceleration of an elevator moving up a shaft. Compare the a(t) curve with the v(t) curve, each point on the a(t) curve shows the slope of the v(t) curve at the corresponding time. When v is constant (at either 0 or 4 m/s), the slope is zero and so also is the acceleration. When the cab first begins to move, the v(t) curve has a positive slope, which means that a(t) is positive. When the cab slows to a stop, the slope of the v(t) curve are negative; that is, a(t) is negative.Next compare the slopes of the v(t) curve during the two acceleration periods. The slope associated with the cab’s slowing down (commonly called “deceleration”) is steeper because the cab stops in half the time it took to get up to speed. The steeper slope means that the magnitude of the deceleration is larger than that of the acceleration, as indicated in Fig. 2-8c.

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Figure 2.9 (a) The x(t) curve for an elevator cab that moves upward along an x axis. (b) The v(t) curve for the cab. Note that it is the slope of the x(t) curve. (c) The a(t) curve for the cab. It is the slope of the v(t) curve.

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2.4 Motion Diagrams

A motion diagram is a representation of a moving object at successive equal time intervals,

Figure 2.12 Motion diagrams of a car moving along a straight roadway in a single direction.

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2.4 Constant Acceleration: A Special Case

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Many applications of mechanics involve objects moving with constant acceleration. This type of motion is important because it applies to numerous objects in nature, such as an object in free fall near Earth’s surface (assuming air resistance can be neglected). A graph of acceleration versus time for motion with constant acceleration is shown in Figure 2.10a.

(a) (b)

(c)

Figure 2.10 A particle moving along the x-axiswith constant acceleration a.(a) the acceleration vs. time graph,(b) the velocity vs. time graph, and(c) the position vs. time graph.

When an object moves with constant acceleration, the instantaneous acceleration at any point in a time interval is equal to the value of the average acceleration over the entire time interval. Consequently, the velocity increases or decreases at the same rate throughout the motion, and a plot of v versus t gives a straight line with either positive, zero, or negative slope.Because the average acceleration equals the instantaneous acceleration when a is constant,

Because acceleration is the slope of the v-t graph, then

Let ti = 0 and tf be any arbitrary time t. Also, let vi = v0 (the initial velocity at t = 0) and vf = v (the velocity at any arbitrary time t). With this notation, we can express the acceleration as

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Equation 2.6 states that the acceleration a steadily changes the initial velocity v0 by an amount at.

The graphical interpretation of v is shown in Figure 2.15b. The velocity varies linearly with time according to Equation 2.6, as it should for constant acceleration.Because the velocity is increasing or decreasing uniformly with time, we can express the average velocity in any time interval as the arithmetic average of the initial velocity v0 and the final velocity v:

We can now use this result along with the defining equation for average velocity, Equation 2.2, to obtain an expression for the displacement of an object as a function of time. Again, we choose ti = 0 and tf = t, and for convenience, we write ∆x = xf - xi = x - x0. This results in

Then

We can obtain another useful expression for displacement by substituting the equation for v (Eq. 2.6) into Equation 2.8:

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This equation can also be written in terms of the position x, since ∆x = x - x0.

Figure 2.15c shows a plot of x versus t for Equation 2.9, which is related to the graph of velocity vs. time: The area under the curve in Figure 2.15b is equal to v0t + (1/2) at2, which is equal to the displacement ∆x. In fact, the area under the graph of v versus t for any object is equal to the displacement ∆x of the object.

Finally, we can obtain an expression that doesn’t contain time by solving Equation 2.6 for t and substituting into Equation 2.8, resulting in

The three most useful equations—Equations 2.6, 2.9, and 2.10—are listed in Table 2.4.

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EXAMPLE 2.4(a) A race car starting from rest accelerates at a constant rate of 5.00 m/s2. What is the velocity of the car after it has traveled 30.5 m? (b) How much time has elapsed? (c) Calculate the average velocity two different ways.

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EXAMPLE 2.4

A car traveling at a constant speed of 24.0 m/s passes a trooper hidden behind a billboard, as in Figure 2.17. One second after the speeding car passes the billboard, the trooper sets off in chase with a constant acceleration of 3.00 m/s2. (a) How long does it take the trooper to overtake the speeding car? (b) How fast is the trooper going at that time?

First: The car:

Since the car is moving at constant velocity of 24 m/s, then its acceleration acar = 0. During 1 second interval, the car has moved a displacement of 24 m as follows:

∆xcar = vcar t = 24 (m/s) x (1 s) = 24 m.

Then at t = 0, the car is at x0car = 24 m and moving at v0car = 24 m/s.

Then, after time t, the car made a displacement ∆xcar given by:

Then, the position of the car after time t is:

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Second: The trooper:

At t = 0, the trooper is at x0trooper = 0 m and starts from rest (v0trooper = 0). Then, the position of the trooper after time t is

The trooper catches up with the car when their positions are the same. Set xtrooper = xcar , and solve the quadratic equation:

ex

A typical jetliner lands at a speed of 71.5 m/s and decelerates at the rate of 4.47 m/s2. If the plane travels at a constant speed of 71.5 m/s for 1.00 s after landing before applying the brakes, what is the total displacement of the aircraft between touchdown on the runway and coming to rest?

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During motion at constant velocity, the a = 0, v0 = 71.5 m/s, and t = 1.00 s, the displacement while the plane is coasting:

Use the time-independent kinematic equation to find the displacement while the plane is braking.

Take a = -4.47 m/s2, v0 = 71.5 m/s and v = 0. The negative sign on a means that the plane is slowing down.

Sum the two results to find the total displacement:

2.6 Freely Falling Objects

When air resistance is negligible, all objects dropped under the influence of gravity near Earth’s surface fall toward Earth with the same constant acceleration. Such motion is called free fall.The expression freely falling object doesn’t necessarily refer to an object dropped from rest. A freely falling object is any object moving freely under the influence of gravity alone, regardless of its initial motion. Objects thrown upward or downward and those released from rest are all considered freely falling.We denote the magnitude of the free-fall acceleration by the symbol g. The value of g decreases with increasing altitude, and varies slightly with latitude as well. At Earth’s surface, the value of g is approximately 9.80 m/s2. Unless stated otherwise, we will use this value for g in doing calculations. For quick estimates, use g = 10 m/s2.If we neglect air resistance and assume that the free-fall acceleration doesn’t vary with altitude over short vertical distances, then the motion of a freely falling object is the same as motion in one dimension under constant acceleration. This means that the kinematics equations developed

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in Section 2.5 can be applied. It’s conventional to define “up” as the + y-direction and to use y as the position variable. In that case the acceleration is a = -g = -9.80 m/s2. In Chapter 7, we study the variation in g with altitude.Example:

A ball is thrown from the top of a building with an initial velocity of 20.0 m/s straight upward, at an initial height of 50.0 m above the ground. The ball just misses the edge of the roof on its way down, as shown in Figure 2.20. Determine (a) the time needed for the ball to reach its maximum height, (b) the maximum height, (c) the time needed for the ball to return to the height from which it was thrown and the velocity of the ball at that instant, (d) the time needed for the ball to reach the ground, and (e) the velocity and position of the ball at t = 5.00 s. Neglect air drag.The diagram in Figure 2.20 establishes a coordinate system with y0 = 0 at the level at which the ball is released from the thrower’s hand, with y positive upward.

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(a) Find the time when the ball reaches its maximum height.

At maximum height, the ball comes to rest and changes its direction of motion, then v = 0 at maximum height. Using

With a = -g = -9.8 m/s2, then:

Then, the time to reach the maximum height is:

(b) Determine the ball’s maximum height.

with y0 = 0, and t = 2.04 s, and using:

(c) Find the time the ball takes to return to its initial position, and find the velocity of the ball at that time.When the ball returns to its initial position, its displacement ∆y = y – y0 = 0. Then:

The ball’s velocity at that instant is:

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(d) Find the time required for the ball to reach the ground.When the ball reaches the ground ( y = -50 m), its displacement ∆y = y – y0 = -50. Then:

Apply the quadratic formula and take the positive root, then:

(e) Find the velocity and position of the ball at t = 5.00 s.

A rocket moves straight upward, starting from rest with an acceleration of +29.4 m/s2. It runsout of fuel at the end of 4.00 s and continues to coast upward, reaching a maximum height before falling back to Earth.(a) Find the rocket’s velocity and position at the end of 4.00 s. (b) Find the maximum height the rocket reaches. (c) Find the velocity the instant before the rocket crashes on the ground.

(a) Phase 1: Find the rocket’s velocity and position after 4.00 s.

Write the velocity and position kinematic equations:

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Substituting a = 29.4 m/s2, v0 = 0, and y0 = 0:

Substitute t = 4.00 s into Equations (3) and (4) to find the rocket’s velocity v and position y at the time of burnout. These will be called vb and yb, respectively.

(b) Phase 2: Find the maximum height the rocket attains.Adapt Equations (1) and (2) to phase 2, substituting a = - 9.8 m/s2, v0 = vb = 118 m/s, and y0 = yb = 235 m:

Substitute v = 0 (the rocket’s velocity at maximum height) in Equation (5) to get the time it takes the rocket to reach its maximum height:

Substitute t = 12.0 s into Equation (6) to find the rocket’s maximum height:

(c) Phase 2: Find the velocity of the rocket just prior to impact.

Find the time to impact by setting y 5 0 in Equation (6) and using the quadratic formula:38

Substitute this value of t into Equation (5):

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If the velocity of a body is plotted against the time, thegraph obtained is a velocity–time graph.

Figure 3.1 Uniform velocity

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Figure 3.2b Non-uniform acceleration

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Figure 3.2a Uniform acceleration

The area under a velocity–time graph measures the distancetravelled.

In Figure 3.1, AB is the velocity–time graph for abody moving with a uniform velocity of 20 m/s.Since distance = average velocity × time, after 5 s itwill have moved 20 m/s × 5 s = 100 m. This is theshaded area under the graph, i.e. rectangle OABC.In Figure 3.2a, PQ is the velocity–time graph for abody moving with uniform acceleration. At the start ofthe timing the velocity is 20 m/s but it increases steadilyto 40 m/s after 5 s. If the distance covered equals thearea under PQ, i.e. the shaded area OPQS, thendistance = area of rectangle OPRS+ area of triangle PQR= OP × OS + 1

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2 × PR × QR(area of a triangle = 12base × height)= 20 m/s × 5 s + 12 × 5 s × 20 m/s= 100 m + 50 m = 150 m

Notes1 When calculating the area from the graph, the unitof time must be the same on both axes.2 This rule for fi nding distances travelled is true evenif the acceleration is not uniform. In Figure 3.2b,the distance travelled equals the shaded area OXY.The slope or gradient of a velocity–time graph represents theacceleration of the body.In Figure 3.1, the slope of AB is zero, as is theacceleration. In Figure 3.2a, the slope of PQ isQR/PR = 20/5 = 4: the acceleration is 4 m/s2.In Figure 3.2b, when the slope along OX changes,so does the acceleration.

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Distance–time graphsA body travelling with uniform velocity coversequal distances in equal times. Its distance–timegraph is a straight line, like OL in Figure 3.3for a velocity of 10 m/s. The slope of the graph isLM/OM = 40 m/4 s = 10 m/s, which is the valueof the velocity. The following statement is true ingeneral:The slope or gradient of a distance–time graph represents thevelocity of the body.

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Both C and D are accelerating but C isaccelerating more than D.

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The change in velocity here is

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8 m/s (not 12 m/s).

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Acceleration doesn’t always haveto be positive.

Lines above the time axis usually show forwardmotion and so lines below the axis show backwards motion.

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Lines above the time axis usually mean forwards motion and linesbelow usually mean backwards motion.

If the line is horizontal then this means that the object istravelling at a constant velocity. The higher the line on thegraph, the higher the velocity. If the object is travellingbackwards, the lower the line, the faster it is travelling.

Both A and B travel at67

constant velocity but A travelsfaster than B.

Both C and D travel atconstant velocity but C travelsfaster than D.

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time/s

distance/m

t1 t2 time/s

d1

d2

distance/m

A= d2 - d1

B= t2 - t1

Distance-Time graphs:

Speed = gradient of the distance-time graph

a) Body at rest (speed = 0)

A horizontal line means that distance is unchanging, i.e. the object is at rest, i.e. the speed is zero.

b) Body moves with constant speed

A straight, inclined line means that the object is moving with constant speed. The gradient of the line equals the speed. To find the gradient:

draw a triangle such as shown above work out A and B from the axes (don't measure them in cm) gradient = A/B

Example:

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time/s

distance/m

t t t

d1

d2

d3

distance/m

d3

Speed = 6 m / 3 s = 2 m/s.

c) Body moves with changing speed

When the distance-time graph is a curve, as shown below, this means that the speed is changing. In Fig. 1. the distance traveled in equal time intervals is increasing (d3 > d2 > d1), so that the speed is increasing with time. The opposite situation is shown in Fig. 1. where the distance traveled in equal time intervals is decreasing (d 3 < d2 < d1), so that the speed is decreasing with time.

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Acceleration

Acceleration is the rate of change of velocity. If acceleration is constant, then:

acceleration = final velocity − initial velocitytime taken

In symbols:

a = v − ut

Units: m/s2 or cm/s2

Example: A train is moving at 40 m/s. Its speed 5 seconds later is 60 m/s. What is the acceleration of the train?

a = v − ut

a =

60 − 405 = 4 m/s2.

Speed-Time graphs:

1- Acceleration is the gradient of the speed-time graph.2- Distance traveled is the area under the speed-time graph.

a) Body at rest (speed = 0)

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time/s

speed /m/s

time/s

speed /m/s

v2

speed/ m/s

The speed of the body is zero for all times.

b) Body moves with constant speed

A horizontal line means that the speed is unchanging, i.e. the object is moving with constant (uniform) speed, i.e. its acceleration is zero.

c) Body moves with constant acceleration

A straight, inclined line means that the object is moving with constant acceleration. The gradient of the line equals the acceleration = A/B.

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time/st1 t2

v1

v2

speed/ m/s

A= v2 - v1

B= t2 - t1

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time/st1 t2

v1

v2

speed/ m/s

B= t2 - t1

The distance traveled by the body between t1 and t2 (time interval t2-t1) equals the shaded area under the speed-time graph shown below.

distance = area = ( v1 + v2

2 )( t2 −t1 )

d) Body moves with changing acceleration

When the speed-time graph is a curve, as shown below, this means that the acceleration is changing. In Fig. 1. the speed is increasing in equal time intervals (v3 > v2 > v1), so that the acceleration is increasing with time. The opposite situation is shown in Fig. 1. where the speed

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time/s

speed/ m/s

t t t

v1

v2

v3

time/s

speed /m/s

t t t

v1

v2

v3

is decreasing in equal time intervals (v3 < v2 < v1), so that the acceleration is decreasing with time.

During a 30-minute round trip to the store, the total distance traveled is 6 km. The average speed is 12 km/h. The displacement for the

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round trip is zero, since there was no net change in position. Thus the average velocity is zero.

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Figure 2.11 Position vs. time, velocity vs. time, and speed vs. time on a trip. Note that the velocity for the return trip is negative.

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