# Measurement Of A Certain Physical Quantity Biology Essay

Published: Last Edited:

This essay has been submitted by a student. This is not an example of the work written by our professional essay writers.

To communicate the result of a measurement of a certain physical quantity, a unit for the quantity must be defined. Imagine you make a curtain and need to buy material. You should know how much material was required, for example, 2 meters wide and 6 meters long. Without the unit the information is incomplete.

It is not just lengths that have units, all physical quantities have units (e.g. mass, time and temperature). Without units much of our work as scientists would be meaningless. We need to express our thoughts clearly and units give meaning to the numbers we calculate. Depending on which units we use, the numbers are different (e.g. 3.8 m and 3800 mm actually represent the same length). Units must be specified when expressing physical quantities.

## 1.1.1 International System of Units

In 1960, an international committee agreed on a set of definitions and standards to describe the physical quantities. The system that was established is called the System International (SI). The System International is built up from three kinds of units: base units, supplementary units and derived units.

## Base Units

There are seven base units for various physical quantities, which is listed in Table 1.1. They are called base units because none of them can be expressed as combinations of the other six.

Table 1.1 Base units

## Symbol

Length

Meter

m

Mass

Kilogram

kg

Time

Second

s

Electric current

Ampere

A

Thermodynamic temperature

Kelvin

K

Intensity of light

Candela

cd

Amount of substance

Mole

mol

## Supplementary Units

Supplementary units have not been classified certain of the SI under either base units or derived units which are introduced in the following. Supplementary units contain two units of purely geometrical quantities, which are plane angle and the solid angle, as shown in Table 1.2.

Table 1.2 Supplementary units

## Symbol

Plane angle

Solid angle

sr

As shown in Figure 1.1, the radian is the plane angle between two radii of a circle which cut off on the circumference an arc, equal in length to the radius. The steradian is the solid angle (three-dimensional angle) subtended at the centre of a sphere by an area of its surface equal to the square of radius of the sphere.

O

r

r

(a)

O

r

r

1sr

(b)

## Derived Units

SI units for measuring all other physical quantities are derived from the base and supplementary units. Some of the derived units are given in Table 1.3

Table 1.3 Derived units

## In terms of base units

Force

Newton

N

Kg m s-2

Work

Joule

J

N m = Kg m2 s-2

Power

Watt

W

J s-1 = Kg m2 s-3

Pressure

Pascal

Pa

N m-2 = Kg m-1 s-2

Electric charge

Coulomb

C

A s

## 1.1.2 The other Systems of Units

The remaining sets of units, such as c.g.s units, imperial units and natural units, are also internationally recognized and still in use by others. We will introduce c.g.s units only.

In the system of c.g.s units, the meter is replaced by the centimeter and the kilogram is replaced by the gram. This is a simple change but it means that all units derived from these two are changed. For example, the units of force and work are different. These units are used most often in astrophysics and atomic physics.

## 1.1.3 How to Convert Units

Sometimes it's necessary to convert units from one set of units to another.

Unit conversions are multiplied and divided just like ordinary algebraic symbols. The most important step of converting units is what we can express the same physical quantity in two different units and form an equality. For example, we consider the case of converting millimeter (mm) to meter (m). We know that 1 m = 1000 mm =103 mm, it mean that 1 meter represents the same physical length as 1000 mm. In physical sense, multiplying by the quantity 1m/1000mm is really multiplying by unity, which does not change the physical meaning of the quantity. Thus, to find the number of meters in 250mm, we write

And to find the millimeters in 0.25m, we write

A ratio of units, such as 1 m = 1000 mm ( or ), is called a conversion factor. Some conversion factors are listed in Table 1.4.

Table 1.4 The conversion factors for length, mass and time.

Length:

1 nanometer = 1 nm = 10-9 m

1 micrometer = 1 µm = 10-6 m

1 millimeter = 1 mm = 10-3 m

1 centimeter = 1 cm = 10-2 m

1 kilometer = 1 km = 103 m

Mass:

1 microgram = 1µg = 10-9 kg

1 milligram = 1 mg = 10-6 kg

1 gram = 1 g = 10-3 kg

1 kilogram = 1 kg = 103 g

Time:

1 nanosecond = 1ns =10-9 s

1 microsecond = 1µs = 10-6 s

1 millisecond = 1ms = 10-3 s

1 minute = 1 min = 60 s

1 hour = 1 h = 60 min = 3600 s

1 day = 24 h = 86400s

When a problem requires calculations using numbers with units, we always write the numbers with the correct units and carry the units through the calculation. This provides a very useful check for calculations. If at some stage in a calculation you find that an equation or an expression has inconsistent units, you know you have made an error somewhere. In this book we will always carry units through all calculations, we strongly urge you follow this practice when you solve problems.

## Example 1.1

If a car is traveling at a speed of 30 m/s, is it exceeding the speed limit of 60 km/h?

## Given

The speed of the car: v = 30 m/s

## Find

Change the speed's unit to km/h.

## Solution

We know that and . Using these conversion factors, we convert meters to kilometers, firstly,

And then convert seconds to hours:

Thus, the driver should slow down because he is exceeding the speed limit.

## 1.2 Scalar and Vector

Some physical quantities, such as time, temperature, mass, density, and electric charge, can be described completely by a single number with a unit. We call such physical quantity a scalar quantity. Calculations with scalar quantities use the operations of ordinary arithmetic. For example, 5 kg + 2 kg = 7 kg or 3.2 m - 2.0 m = 1.2 m.

Many other quantities, such as displacement, velocity, acceleration, force, have both a magnitude and a direction in space. We call such physical quantity a vector quantity. A vector is represented graphically by a directed line segment with an arrowhead. The length of the line segment, according to a chosen scale, corresponds to the magnitude of vectors, and the direction of the arrow gives the direction of the vector.

We usually represent a vector quantity by a single letter, such as force , velocity and displacement . In this book, we always print vector symbols in boldface type. In handwriting, vector symbols are usually written with an arrow above to indicate that they represent vector quantities, such as . If we wish to refer only to the magnitude of a vector , we use light face type such as .

Vector quantities play an essential role in all areas of physics. Vectors are mathematical objects and we use them to describe physics in the language of mathematics. Now let's talk about what vectors are and how they combine.

We define two force vectors. is 3 N in the forward direction and is 4 N in the upward direction, which are shown in Figure 1.2 (a). We call force the vector sum of forces and . This relationship is symbolically expressed as

The final answer when adding vectors is called the resultant. There are two primary graphical techniques to get the sum of vectors, one is the tail-to-head method and another is the parallelogram method.

As shown in Figure 1.2 (b), we add the second vector at the end of the first vector The vector from the tail of the first vector (the starting point) to the head of the last (the end point) is then the sum of the vectors. This is the tail-to-head method of vector addition.

As shown in Figure 1.2 (c), when vectors and are both drawn with their tails at the same point, the sum of vectors is the diagonal of a parallelogram constructed with and as two adjacent sides. This is the parallelogram method.

(a)

(b)

(c)

## Example 1.2

If you walk 400 m north, then 300 m east, how far from the starting point are you?

(A) 100 m; (B) 500 m; (C) 700 m.

Figure 1.3 Example 1.3.

(North); (east).

## Find

The sum vector of and .

## Solution

The two legs of your trip form the sides of a right triangle; your final displacement is the hypotenuse. As shown in Figure 1.3, we can calculate the length of the vector sum, 500 m.

## Example 1.3

(1) The sum of two parallel vectors.

## =

(2) The sum of two antiparallel vectors.

## =

(3) The sum of any two vectors.

## Example 1.4

As shown in Figure 1.4 (a), what is the sum of vectors?

## Given

Four vectors as show in Figure 1.4 (a).

The sum vector.

## Solution

From the tail-to-head method, the resultant vector , as shown in Figure 1.4 (b) is the sum of vectors.

(b)

(a)

Figure 1.4 (a) Vectors A, B, C and D;

(b) The resultant vector E is the sum of vectors.

## 1.2.2 Subtraction of Vectors

is a vector having the same magnitude as but opposite direction. We define the difference of two vectors and to be the vector sum of and , namely,

Therefore, subtracting a vector from another is the same as adding the vector in the opposite direction. An example is given in Figure 1.5.

Figure 1.5 Subtracting from is equivalent to adding to .

Therefore, the sum of a vector and its negative vector is a null vector. Null vector is a vector of zero magnitude and arbitrary direction.

## 1.2.3 Components of a Vector

A component of a vector is its effective value in a given direction. A vector may be considered as the resultant of its component vectors along the specified direction. It is usually convenient to resolve a vector into components along mutually perpendicular directions. Such components are called rectangular components.

Consider a vector in a rectangular coordinate system, as shown in Figure 1.6. It can be expressed as the sum of two vectors: , parallel to the x-axis; and , parallel to the y -axis. Mathematically,

## ,

where and are the component vectors of . The projection of along the x-axis, , is called the x-component of , and the projection of along the y-axis, , is called the y-component of . These components can be either positive or negative numbers with units. From the definitions of sine and cosine, the components of are

(1.1)

These components form two sides of a right triangle having a hypotenuse with magnitude A. It follows that 's magnitude and direction are related to its components through the Pythagorean theorem and the definition of the tangent:

(1.2)

This formula gives the right answer only half the time. The inverse tangent function returns values only from -90o to 90o, so the answer in your calculator window will only be correct if the vector happens to lie in first or fourth quadrant. If it lies in second or third quadrant, adding 180o to the number in the calculator window will then give the right answer.

x

y

O

x

y

O

Figure 1.6 Any vector lying in the xy-plane can be represented by its rectangular components Ax and Ay .

## Example 1.5

In Figure 1.7, there are two vectors lying in the xy-plane, where , , , . Determine the x- and y-components of , and .

## Given

Vector : A = 10, ;

Vector : B = 6, .

## Find

The components of , and .

## Solution

Use Eq. 1.1 to find the components of :

The components of :

Because of , the components of :

Find the magnitude and direction of :

## O

Figure 1.7 (Example 1.5)

Therefore, the magnitude of is 11.21, and its direction is 32.7o with x-axis.

## 1.2.4 Product of Two Vectors

There are two types of vector multiplications. The products of these two types are known as scalar product and vector product. As the name implies, scalar product of two vector quantities is a scalar quantity, while vector product of two vector quantities is a vector quantities.

## Scalar or Dot Product

The scalar product of two vectors and is written as , and defined as

(1.3)

Where and are the magnitudes of vector and , and is the angle between them.

Since and , hence, . The order of multiplication is irrelevant. In other words, scalar product is commutative.

## Example 1.6

(1) Give the scalar product of two parallel vectors.

Give the scalar product of two mutually perpendicular vectors.

(1) Since,

(2) Since,

## Remarks

(1) The scalar product of two parallel vectors is equal to the product of their magnitudes.

(2) The scalar product of two mutually perpendicular vectors is zero.

## Example 1.7

(1)

(2) Give the scalar product of two antiparallel vectors.

## Solution

(1)

(2) Since and the angle between and is 180o,

## Vector or Cross Product

The vector product of two vectors and , is a vector which is written as . Its magnitude is, where is the angle between and . Its direction can be determined by right hand rule. For that purpose, as shown in Figure 1.8, place together the tails of vectors and to define the plane of vectors and . The direction of the product vector is perpendicular to this plane. Rotate the first vector into through the smaller of two possible angles and curl the fingers of the right hand in the direction of rotation, keeping the thumb erect. The direction of the product vector will be along the erect thumb.

Figure 1.8 The right hand rule.

In summary, we can write the vector product of two vectors as follows

(1.4)

where is the magnitude, and is the direction. is perpendicular to the plane containing and given by right hand rule for the vector product of two vectors.

Because of this direction rule, is a vector opposite in sign to . Hence,

. (1.5)

The cross product is non commutative.

## Example 1.8

(1) Give the cross product of two parallel vectors.

(2) Give the magnitude of the cross product of two perpendicular vectors.

## Solution

(1) Since , the cross product of such two parallel vectors is .

(2) Since , the magnitude of the cross product of two perpendicular vectors is .

## Remarks

(1) The cross product of two parallel vectors is null vector.

(2) The magnitude of the cross product of two perpendicular vectors is .

## 1.3 Motion

Everything in the vastness of space is in a state of perpetual motion. We move around the Earth's surface, while the Earth moves in its orbit around the Sun. The Sun and the stars are in motion too. In every piece of matter, the atoms are in a state of never ending motion.

The branch of physics concerned with the study of the motion of an object and the relationship of this motion to such physical concepts 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, we shall focus on kinematics and on one- dimensional motion, namely, rectilinear motion. Rectilinear motion means motion along a straight line.

## 1.3.1 Displacement

Whenever a body moves from one position to another, the change in its position is called displacement. The displacement can be represented as a vector that describes how far and in what direction the body has been displaced from its original position.

As shown in Figure 1.9 (a), is the position vector of and is the position vector of . Then the displacement is a change on the position of body from its initial position A to its final position B. The magnitude of is the straight line distance between the initial and final positions of the body. When the body moves along a straight line, the displacement coincides with the path of motion as shown in Figure 1.9 (b).

x

y

A

B

O

(b)

x

y

A

B

O

(a)

s

Figure 1.9 An object moving along some curved path between points A and B. The displacement vector is the difference in the position vectors: .

## 1.3.2 Velocity

The time rate of change of displacement is known as velocity. Its direction is along the direction of displacement. So if is the total displacement of body in time t, and then its average velocity during the interval t is defined as

(1.6)

Because the path may be straight or curved and the motion may be steady or variable, average velocity does not tell us everything about the motion. In such case the motion is described by the instantaneous velocity. At any time t, let the body be at point A in Figure 1.9 (a), its position is given by position vector . After a short time interval following the instant t, the body reaches the point B which is described by the position vector . The displacement of the body during this short time interval is given by

(1.7)

The notation (delta) is used to represent a very small change. The instantaneous velocity at point A, can be found by making smaller and smaller. In this case will also become smaller and point B will approach A. Therefore, the instantaneous velocity is defined as the limiting value of as approaches zero, namely,

(1.8)

If the instantaneous velocity does not change, the body is said to be moving with uniform velocity.

In day-to-day usage, the terms speed and velocity are interchangeable. In physics, however, there is a clear distinction between these two quantities. The average speed of an object, a scalar quantity, is defined as the total distance traveled divided by the total time. For example, in Figure 1.9 (a), supposed an object moves from point A to B during a time interval t. The average velocity of the object is Eq. (1.6), while its average speed is

(1.9)

When the body moves along a straight line, as shown in Figure 1.18(b), the magnitude of the average velocity and the average speed are the same.

The SI unit of average speed is the same as the unit of average velocity: meters per second. However, unlike average velocity, average speed has no direction and hence carries no algebraic sign.

## Example 1.9

A cyclist moves from A through B to C in 10 seconds as shown in Figure 1.10. Calculate both his average speed and his average velocity.

## Given

A cyclist moves from A through B to C in 10 seconds;

(right); (upward).

## Find

The average speed and the average velocity.

Figure 1.10 A cyclist moves from A through B to C in 10 seconds.

## Solution

The speed of cyclist is a scalar, and it is

The velocity of cyclist is a vector. Firstly we should find the displacement of the cyclist. Its displacement is

The magnitude of the displacement is

## ,

and its direction is from A to C. Therefore, the average velocity of cyclist is

in direction from A to C.

## 1.3.3 Acceleration

If the velocity of a body changes, it is said to be moving with an acceleration. The time rate of change of velocity of a body is called acceleration.

As velocity is a vector so any change in velocity may be due to a change in its magnitude or in its direction.

As shown in Figure 1.11, consider a body whose velocity at any instant t changes to in further small time interval . The change in velocity is . The average acceleration during time interval is given by

(1.10)

Figure 1.11 The velocity vector is the difference in the position vectors: .

The instantaneous acceleration is

(1.11)

If the velocity of a body is increasing, its acceleration is positive but if the velocity is decreasing the acceleration is negative. For a body moving with uniform acceleration, its average acceleration is equal to instantaneous acceleration.

## 1.3.4 Motion graph

Graph is an important tool for picturing certain concepts. Below are some graphs that help us picture the concepts of displacement, velocity and acceleration.

## Position-Time Graph

The position-time graph may be used to illustrate the variation of position of a body with time. In Figure 1.12 there are three examples of position-time graphs.

Position

Position

. A

t

Position

Figure 1.12 Some common position-time graphs

We know that the gradient (slope) of a graph is defined as the change in y divided by the change in x, i.e . In the position-time graph the gradient of the graph is , and this is just the expression for velocity. Therefore, the slope of position-time gives the velocity.

In Figure 1.12, (a) shows the graph for an object stationary over a period of time. The gradient is zero, so the object has zero velocity. (b) shows the graph for an object moving at a constant velocity. The displacement is increasing as time goes on. The gradient stays constant, so the velocity is constant. Here the gradient is positive, so the object is moving in the direction we have defined as positive. (c) shows the graph for an object moving at a constant acceleration. The gradient is increasing with time, thus the velocity is increasing with time and the object is accelerating. The point A in graph corresponds to time t. The magnitude of the instantaneous velocity at this instant is numerically equal to the slope of the tangent at the point A.

## Example 1.10

A person runs 3.0 km in 30min with a constant velocity, and then stops for a 30 min nap. Upon awakening, he runs twice as fast and finishes his running in a total time of 90 min. (a) Calculate the speed before he stopped for a nap. (b) Draw the position-time graph. (c) Calculate the average speed of the person.

## Given

In the first 30 min, the person runs 3.0km. In the second 30 min, the person stops for a nap. In the last 25min, the person run twice as fast.

## Find

(a) The average speed in the first 30 min

(b) Draw the position-time graph of the person runs;

(c) The average speed of the runner.

## Solution

(a) According to the meaning of the problem, the average speed before the person stopped for a nap is

(b) In the first 30 min, the average speed is . In the second 30 min, the person stops for a nap, so the speed is . In the last 25min, the speed is , so in the last 30 min, he runs

Thus, the position-time graph is as shown in Figure 1.13.

A

B

C

30

60

0

t (s)

x(km)

9

90

3

6

Figure 1.13 The position-time graph for the runner.

(c) In the last 30 min, the distance of the person runs is 6 km. Therefore, the person runs 9 km in 90 min. The average speed is

## Velocity-Time Graph

The velocity-time graph may be used to illustrate the variation of velocity of a body with time. The velocity-time graphs of an object making three different journeys along a straight road are shown in Figure 1.13. In the velocity-time graph the gradient of the graph is , and this is just the expression for acceleration. Therefore, the slope of velocity-time gives the acceleration.

velocity

velocity

velocity

t

## v

t

. A

t

Figure 1.14 Some common velocity-time graphs

As shown in Figure 1.14, (a) shows the graph for an object moving at a constant velocity. When the velocity of an object is constant, its velocity-time graph is a horizontal straight line. (b) shows the graph for an object moving at a constant acceleration. The velocity-time graph is a straight line. The gradient stays constant, so the acceleration is constant. (c) shows the graph for an object moving with increasing acceleration. The velocity-time graph is a curve. The point A in graph corresponds to time t. The magnitude of the instantaneous acceleration at this instant is numerically equal to the slope of the tangent at the point A.

The distance moved by an object can also be determined by using its velocity-time graph. The area between the velocity-time graph and the time axis is numerically equal to the distance covered by the object. For one example, when an object moves at constant velocity for time t as shown in Figure 1.14(a), the distance covered by the object given by Eq. (1.5) is . This distance can also be found by calculating the area under the velocity-time graph, and the area is shown shaded in Figure 1.14(a). For another example, in Figure 1.14(b), the velocity of the object increases uniformly from 0 to in time t. The magnitude of its average velocity is give by

Then, the distance covered is

## ,

which is equal to the area of the triangle shaded in Figure 1.14(b).

## Example 1.11

The velocity-time graph of a body moving on a straight path is shown in Figure 1.15. Describe the motion of the body and find the distance covered.

20

30

0

t (s)

v (m/s)

10

A

B

C

B'

A'

D'

D

35

-5

Figure 1.15 The velocity-time graph of a body moving from A through B to C.

## Given

The velocity-time graph of a body moving.

## Find

The displacement and the distance.

## Solution

The graph tell us that the velocity of the body remains 10m/s from 0 to 20th second, and then decreases uniformly to -5m/s from 20th to 35th second. The acceleration of the body during the last 15 seconds is

The negative sign indicates that the velocity of the body decreases during the last 15 seconds.

The distance covered by the body is equal to the area between the velocity-time graph and the time-axis. Thus, the displacement of the body travelled is

The distance of the body travelled is

## Acceleration-time graph

In this chapter on rectilinear motion we will only deal with objects moving at a constant acceleration. Two kinds of acceleration-time graphs are shown in Figure 1.16.

10s

2 m/s2

Figure l.16 Two kinds of acceleration-time graph.

Figure 1.16 (a) shows the graph for an object which is either stationary or traveling at a constant velocity. Either way, the acceleration is zero over time. (b) shows the graph for an object moving at a constant acceleration. In this case the acceleration is positive (remember that it can also be negative). We can obtain the velocity of a particle at some given time from an acceleration-time graph. It is just given by the area between the graph and the time-axis. For example, in Figure 1.16 (b), showing an object at a constant positive acceleration, the increase in velocity of the object after 10 seconds is the area of the shad, namely,

Figure 1.17 shows how displacement, velocity and time relate to each other. Given a displacement-time graph like (a), we can plot the corresponding velocity-time graph by remembering that the slope of a position-time graph gives the velocity. Similarly, we can plot an acceleration-time graph from the gradient of the velocity-time graph.

time

time

time

(a)

(b)

(c)

Figure 1.17 A Relationship Between Displacement, Velocity and Acceleration.

## 1.4 Uniformly accelerated motion

Uniformly accelerated motion is important because it applies to many objects in nature. For example, an object in free fall near Earth's surface moves in the vertical direction with constant acceleration, assuming that air resistance can be neglected. 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 versus t gives a straight line with either positive or negative slope.

As shown in Figure1.18, suppose an object is moving with uniform acceleration along a straight line. If its initial velocity is , and final velocity after a time interval t is , the acceleration is as follows:

Slope = a

vt

vt

Figure 1.18 The velocity-time graph for a Uniformly accelerated motion.

or (1.12)

Because the velocity is increasing uniformly with time, we can express the average velocity in any time interval as the arithmetic average of the initial velocity and the final velocity :

(1.13)

Please note that this expression is valid only when the acceleration is constant, in which case the velocity increases uniformly. Let the distance covered during this time interval be S, then we have

(1.14)

In fact, the area under velocity-time graph for any object is equal to the displacement of the object. Thus, according to Figure 1.18, we can also obtain Eq. (1.14). Finally, we can obtain an expression that does not contain time by solving Eq. (1.12) for t and substituting into Eq. (1.14), resulting in

(1.15)

These equations are useful only for linear motion with uniform acceleration. When the object moves along a straight line, the direction of motion does not change. In this case all the vectors can be manipulated like scalars. In such problem, the direction of initial velocity is taken as positive. A negative sign is assigned to quantities when direction is opposite to that of initial velocity.

The equations for uniformly accelerated motion can also be applied to free fall motion of the objects by replacing by . is the acceleration due to gravity, and its average value near the Earth surface is taken as 9.8 m/s2 in the downward direction.

## (1.1 Units)

1. The speed of light is about . Convert this to kilometer per hour.

## (1.2 Scalar and Vector)

2. The magnitude of two vectors: A = 15 m and B = 3 m. The largest and smallest possible values for the magnitude of the resultant vector are (a) 14.4 and 4, (b) 12 and 8, (c) 18 and 12, (d) none of these answers.

3. If vector B is added to vector A, the resultant vector A + B has magnitude A - B when A and B are (a) perpendicular to each other, (b) oriented in the dame direction, (c) oriented in opposite directions, or (d) in any direction relative to each other.

## (1.3 Motion)

4. Let us define eastward as negative and westward as positive. The correct conclusion is (a) If a car is traveling eastward, its acceleration must be eastward. (b) If the car is slowing down, its acceleration may be positive. (c) A particle with constant nonzero acceleration can never stop and stay stopped.

5. As shown in Figure 1.19, (a), (b) and (c) represent three graphs of the velocities of different objects moving in straight-line paths as functions of time. The possible accelerations of each object as functions of time are shown in (d), (e) and (f). Math each velocity-time graph with the acceleration-time graph that best describes the motion.

Figure 1.19 (Exercise 4) Math each velocity-time graph to its acceleration-time graph.

6. The three graphs in Figure 1.20 represent the position versus time for objects moving along the x axis. Which of these graphs is not physically possible?

Figure 1.20 Which graph is not physically possible?

7. A ball is thrown vertically upward. While the ball is in free fall, does its acceleration (a) increase, (b) decrease, or (c) remain constant?

8. A person travels by car from one city to another with different constant speeds. He drives for 20.0 min at 60 km/h, 15min at 50km/h, and 35 min at 80 km/h. (a) Determine the average speed for the trip. (b) Determine the distance between the initial and final cities along this route.

9. An object moves along the x axis, and its position-time graph is shown in Figure 1.21. Find the average velocity in the time intervals (a) 0 to 1.00 s, (b) 0 to 2.00 s, (c) 2.00 s to 4.00 s, (d) 0 to 5.00 s. Find the instantaneous velocity at the instants (a) t = 0.5 s, (b) t = 2.0 s, (c) t = 3 s, (d) t = 4.5 s.

Figure 1.21 (Exercise 9)

10. The velocity-time graph for an object moving along a straight path is shown in Figure 1.22. (a) Find the average accelerations of this object during the time intervals 0 to 5.0 s, 5.0 s to 15 s, and 0 to 20 s. (b) Find the instantaneous acceleration at 2.0 s, 10 s and 18 s.

Figure 1.22 (Exercise 10)

## (1.4 Uniformly Accelerated Motion)

11. After a ball is thrown vertically upward and is in air, its speed (a) increases, (b) decreases, (c) decreases and then increases, or (d) remains the same.

12. A certain car is capable of accelerating at a rate of . How long does it take for this car to go from a speed of 55 m/s to a speed of 60 m/s.

13. An object starts from rest and accelerates at for the entire distance of 400m. (a) How long did it take the object to travel this distance? (b) What is the speed of the object at the end of the run?

14. An object accelerates uniformly from rest to a speed of 40m/s in 12.0s. (a) Find the distance the object travels during this time, (b) the constant acceleration of the object.