Hafez a radi, john o rasmussen auth principles of physics for scientists and engineers 03

3.2 Average Velocity and Average Speed Average Velocity Consider a particle moving along the x-axis, where its position-time graph is as shown in Fig.3.2.. The slope of the line PQ measu

34 2 Vectors(2) A car travels 6 km due east and then 4 km in a direction 120◦north of east Useboth the graphical and analytical methods to find the magnitude and direction

of the car’s displacement vector

(3) Vector →

A has a magnitude of 10 units and makes 60◦with the positive x-axis.Vector →

B has a magnitude of 5 units and is directed along the negative x-axis.

Use geometry to find: (a) the vector sum→

A +B→, and (b) the vector difference

→

A −B→

(4) A car travels in a circular path of radius 10 m (a) If the car traveled one half ofthe circle, find the magnitude of the displacement vector and find how far thecar traveled (b) Answer part (a) if the car makes one complete revolution

Section 2.3 Vector Components and Unit Vectors

(5) Vector →

A has x and y components of 4 cm and−5cm, respectively Vector B→

has x and y components of−2 cm and 1cm, respectively If →A −B→+ 3C→= 0,

then what are the components of →

C.(6) Three vectors are oriented as shown in Fig.2.15, where A = 10, B = 20, and

C = 15 units Find: (a) the x and y components of the resultant vector D→=A→+

→

B +C→, (b) the magnitude and direction of the resultant vector.

Fig 2.15 See Exercise (6)

(7) The radar beam of a police car points at an angle of 30◦away from the direction

of a highway The radar records the component of the car’s speed along thebeam asv CR = 120 km/h, see Fig.2.16 (a) What is the speedv C of the caralong the highway? (b) Can the radar beam be directed perpendicular to thedirection of the highway? Why or why not?

(8) A radar device detects a rocket approaching directly from east due west Atone instant, the rocket was observed 10 km away and making an angle of 30◦above the horizon At another instant the rocket was observed at an angle of 150◦

in the vertical east-west plane while the rocket was 8 km away, see Fig.2.17.Find the displacement of the rocket during the period of observation.

Fig 2.17 See Exercise (8)

(9) Find the vector components of the sum →

R of the displacement vectors →

A

and →

B whose components along three perpendicular directions are A x = 2,

A y = 1, A z = 3, B x = 1, B y = 4, and B z = 2 Find the magnitude of R→.

(10) Two vectors →

B (of lengths A and B , respectively) make an angle θ

with each other when they are placed tail to tail, see Fig.2.18 (a) By takingcomponents along two perpendicular axes, prove that the length of their vectorsum→

R =→A +B→is:

R=A2+ B2+ 2AB cos θ

(b) For the difference→

C =→A −B→, where C is the length of the third side of a

triangle formed from connecting the head of→

B to the head of →

A as in Fig.2.19,use the same approach to prove that:

(11) A position vector→r = x→i + y→j + z→k makes anglesα, β, and γ with the

x, y, and z axes of a perpendicular right-handed coordinate system as in

Fig.2.20 Show that the relation between what is known as the direction cosinescosα, cos β, and cos γ are as follows: cos2α + cos2β + cos2γ = 1.

Fig 2.20 See Exercise (11)

(13) Two vectors are given by →

A = 2→i + 3→j and→

B = 4→i − 3→j Find: (a) themagnitude and direction of the vector sum →

R =→A +B→, (b) the magnitude and

direction of the vector difference →

S =A→−→B.(14) Two vectors are given by→

A = −→i +→j + 4→k and→

B = 3→i − 4→j +→k Find:(a)→

A +B→, (b) A→−→B , and (c) a vector C→such that→

B by using vector components

(16) Show that for any vector →

A: (a) A→• →

A = A2and (b) →

A×A→= 0

(17) In Exercise 10, show that dotting vector →

R with itself and dotting vector →

C

with itself leads directly to the results of both part (a) and part (b)

(18) For the vectors in Fig.2.21, find the following: (a) →

4 5

x y

(19) (a) Show that →

A • (→A×B→) = 0 for all vectors →A and →

B make an acute angleθ with each other when they are

placed tail to tail as shown in Fig.2.22 (a) Prove that the area of the trianglethat is contained by these two vectors is12|→A×→B| (b) Show that the area ofthe parallelogram formed by→

B is|A→×→B|

(21) Show that →

A • (→B×C→) is equal in magnitude to the volume of the

paral-lelepiped whose sides are formed from the three vectors →

A ,→B , and C→as shown

in Fig.2.23

(22) In the xy plane, point P has coordinates (x1, y1) and is described by the

posi-tion vector →r

1 = x1→i + y1→j Similarly, point Q has coordinates (x2, y2)

and is described by the position vector→r

F = q(→v × B→) gives the force F→on an electric point charge q

moving with velocity→v through a uniform magnetic field →B Find the force on

a proton of q = 1.6×10−19coulomb moving with velocity→v = (2→i +3→j +

Mechanics

Motion in One Dimension 3

Mechanics is the science that deals with motion of objects It is basic to all other

branches of physics The branch of mechanics that describes the motion of objects

is called kinematics In this branch we answer questions like “Does the object speed

up, slow down, stop, or reverse direction?” and “How is time involved in thesesituations?”

In this chapter, we only study motion along straight lines The moving object of

concern is either a particle (a point-like object) or an object that can be viewed to

move like a particle

3.1 Position and Displacement

To locate an object in one-dimensional space, we find its position with respect to

some reference point, called the origin of an axis, such as the x-axis shown in Fig.3.1

The positive/negative direction of this axis is the direction of increasing/decreasing

numbers

A change in the object’s position from an initial position xi to a final position xf

is called displacementx (read delta x), where:

-1-2-3

Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-23026-4_3,

© Springer-Verlag Berlin Heidelberg 2013

x = xf− xi (3.1)

The displacement is a vector quantity which has a magnitude and a direction The

magnitude is the distance between the initial and final positions and the direction isrepresented in Fig.3.1by a plus or minus sign for motion to the right or to the left,respectively

3.2 Average Velocity and Average Speed

Average Velocity

Consider a particle moving along the x-axis, where its position-time graph is as

shown in Fig.3.2 At point P, let its position be xi when the time was tiand at point

Q, let its position be xf when the time was tf (the indices i and f refer to the initialand final values for the variables under consideration) Accordingly, during the timeintervalt = tf− ti , the particle’s displacement is x = xf− xi

Fig 3.2 The position-time graph for a particle moving along the x-axis The slope of the line PQ measures

the average velocityv

One of several quantities associated with the phrase “how fast” a particle moves

is the average velocity,v, which is defined as follows:

Average velocity

The average velocity,v, of a particle is defined as the ratio of its displacement,

x, to the time interval, t That is:

3.2 Average Velocity and Average Speed 43

v = x

t =

xf− xi

From this definition,v has the dimension of length divided by time, that is m/s

in SI units The average velocity is a vector quantity which has a magnitude anddirection represented by a plus or minus sign for motion to the right or to the left,respectively, see Fig.3.1

A car moving along the x-axis starts from the position xi = 2 m when ti= 0 and

stops at xf = − 3 m when tf = 2 s (a) Find the displacement, the average velocity,

and the average speed during this interval of time (b) If the car goes backward

and takes 3 s to reach the starting point, then repeat part (a) for the whole time

interval

Solution: (a) The car’s displacement, see Fig.3.3, is given by:

x = xf− xi = −3 m − 2 m = −5 mThe average velocity is then given by:

Since x and v are negative for this time interval, then the car has moved to the left, toward decreasing values of x, see Fig.3.3 The total covered distance is

d = 5 m and the average speed is thus:

(b) After the backward movement, the final position and final time of the car

are xf = 2 m and tf = 2 s + 3 s = 5 s, respectively, while the total distance covered by the car is d = 5 m + 5 m = 10 m As we know, the displacement

involves only the initial and final positions and will be:

x = xf− xi= 2 m − 2 m = 0Then, the average velocity will be:

3.3 Instantaneous Velocity and Speed

More commonly, we ask how fast a particle is moving at a given instant, which

refers to its instantaneous velocity (or simply velocity) The velocity at any instant

is obtained from the average velocity by allowing the time intervalt to approach

zero Consider the motion of an object (for example a car) This object can be viewed

3.3 Instantaneous Velocity and Speed 45

as a particle for simplicity The motion of that particle between two points P and

Q on a position-time graph is shown in the right part of Fig.3.4 As point Q is brought closer and closer to point P (through points Q1 , Q2, ), the time intervals (t1, t2, ) get progressively smaller The average velocity for each time interval

is the slope of the dotted line in Fig.3.4 As point Q approaches P, the time interval

approaches zero, while the slope of the dotted line approaches the slope of the tangent

to the curve at point P This slope is defined to be the instantaneous velocity v at the time ti In short, we define:

Instantaneous velocity

The instantaneous velocity,v, of a particle is defined as the limiting value of the

ratiox/t as t approaches zero Mathematically v can be expressed as:

velocityv for the interval PQ approaches the slope of the tangent line at P, which is defined as the

instantaneous velocityv at point P

In calculus notation, the above limit is called the derivative of x with respect to t, and written as dx /dt (abbreviated as ˙x) Thus:

The instantaneous velocity,v, can be positive, negative, or zero, depending on the

slope of the position-time graph at the interval of interest in Fig.3.5 In this figure,

v = 0 represents the turning point, and occurs at any maximum or minimum of the x-t graph From here on, we use the word velocity to denote instantaneous velocity.

The speed of a particle is defined as the magnitude of its velocity.

Fig 3.5 The position-time graph for a particle moving along the x-axis On this graph we display:

(1) Positive velocities, where the slope of the tangent lines are positive, (2) Negative velocities, where the slope of the tangent lines are negative, (3) Zero velocities (turning points), where the slope of the tangent

lines are zero, and (4) Inflection points at t1and t2, where the increase/decrease of the velocity reaches a

maximum/minimum

Example 3.2

A particle moves along the x-axis and its coordinates vary with time according

to the relation x = t2− 2 t, where x is measured in meters and t is in seconds.

The position-time graph for this motion is shown in Fig.3.6 (a) Use this graph tocomment about the particle’s motion (b) Find the displacement and the averagevelocity of the particle in the time intervals 0 ≤ t ≤ 1 s and 1 s ≤ t ≤ 3 s (c) Find the velocity of the particle at t = 2 s.

Solution: (a) The particle starts from the origin of the x-axis and moves in the

negative x direction for the first second Its velocity is zero at x = −1 m when

t = 1 s and then heads back in the positive x direction for t > 1 s.

(b) In the interval 0≤ t ≤ 1 s we have ti = 0 and tf = 1 s Since x = t2− 2 t,

we get xi = t2

i − 2 ti = 0 and xf = t2

f − 2 tf = −1 m Thus:

x = xf− xi = −1 m − 0 m = −1 m

3.3 Instantaneous Velocity and Speed 47

Fig 3.6

-10123

In the interval 1 s≤ t ≤ 3 s we have ti = 1 s and tf = 3 s Again, from x =

(c) To find the instantaneous velocity at any time t, we use Eq.3.5and apply

the rules of differential calculus on the coordinate x = t2− 2 t That is:

v = dx

dt = d (t2− 2t)

dt = 2 t − 2

Notice that this expression gives the velocityv at any time t and indicates that v

is increasing linearly with time It tells us thatv < 0 during the interval 0 ≤ t < 1 s (i.e the particle is moving in the negative x direction), and that v = 0 at t = 1 s,

and finallyv > 0 for t > 1 s When t = 2 s we use the above expression to get:

v = 2 × 2 − 2 = 2 m/s

3.4 Acceleration

When the velocity of a particle changes with time, the particle is said to be

acceler-ating Consider the motion of a particle along the x-axis If the particle has a velocity

viat time tiand a velocityvfat time tfas in the velocity-time graph of Fig.3.7, then

we define the average acceleration as:

Average acceleration

The average acceleration, a , of a particle is defined as the ratio of the change

in velocityv = vf− vito the time intervalt = tf− ti That is:

a= v t =vf− vi

a t

Δ

=Δ

Fig 3.7 The velocity-time graph for a car (or simply a particle) moving in a straight line The slope of

the straight line PQ is defined as the average acceleration in the time interval t = t − t

3.4 Acceleration 49Acceleration is a vector quantity having dimensions of length divided by(time)2, or

L/T2; that is m/s2in SI units

It is useful to define the instantaneous acceleration as the limit of the averageacceleration whent approaches zero Consider the motion of a particle (for example

a car that moves like a particle) between the two points P and Q on the

velocity-time graph shown in the right part of Fig.3.8 As point Q is brought closer and closer to point P (through points Q1 , Q2, ), the time intervals (t1, t2, ) get

progressively smaller The average acceleration for each time interval is the slope ofthe dotted line in Fig.3.8 As Q approaches P, the time interval approaches zero, while

the slope of the dotted line approaches the slope of the tangent to the curve at point

P The slope of the tangent line to the curve at P is defined to be the instantaneous

acceleration a at the time ti That is, we define the following:

Instantaneous acceleration

The instantaneous acceleration, a, of a particle is defined as the limiting

value of the ratio v/t when t approaches zero Mathematically a can

In calculus notation, the above limit is called the first derivative of v with respect

to t, and written as d v/dt (simplified sometimes as ˙v), or the second derivative of x with respect to t, and written as d2x /dt2(simplified sometimes as¨x) Thus:

a= d v

dt =d2x

dt2 ≡ vf− vi=tf

ti

a dt ≡ Area under a-t graph (3.8)

From here on, we use the word acceleration to designate instantaneous

acceler-ation Depending on the slope of the tangent to the velocity-time graph, acceleration

a can be positive, negative (called deceleration), or zero If a= 0 for a specific timeinterval in thev − t graph, then the velocity must be a constant in this interval.

(c) To describe the particle’s motion for t≥ 0 we examine the expressions

x = t3− 12 t + 20, v = 3 t2− 12, and a = 6 t.

At t = 0, the particle is at x = 20 m from the origin and moving to the left

with velocityv = −12 m/s and not accelerating since a = 0, see Fig.3.9

3.4 Acceleration 51

At 0< t < 2 s, the particle continues to move to the left (x decreases), but at a decreasing speed, because it is now accelerating to the right, a= positive (Check

the expressions of x, v, and a for t = 1 s and compare the results with Fig.3.9)

At t = 2 s, the particle stops momentarily (v = 0) to reverse its direction of motion At this moment x = 4 m, i.e it will be as close as it will ever be to the

origin It will continue to accelerate to the right at an increasing rate, see Fig.3.9

For t > 2 s, the particle continues to accelerate and move to the right, and its

velocity, which is now to the right, increases rapidly, see Fig.3.9

Fig 3.9

0510152025

-20-1001020

X Data

05101520