(1) read sec. 2-10. (2) 1-6, 16, 20, 29-31, 33, 46, 48,...

Homework:

(1) Read Sec. 2-10.

(2) 1-6, 16, 20, 29-31, 33, 46, 48, 50

(a) the average velocity:

As you walk along a straight line: (b) (c)

2. Compute your average velocity in the following two cases: (a) You walk 73.2 m at a speed of 1.22 m/s and then run 73.2 m at a speed of 2.85 m/s along a straight track. (b) You walk for 1.0 min at a speed 1.22 m/s and then run for 1.0 min at 3.05 m/s along a straight track. (c) Graph x versus t for both cases and indicate how the average velocity is found on the graph.

interval timentdisplaceme

=avgv

)m/s(71.1

2.8573.2

1.2273.2

73.273.2=

+

+=avgv

)/(14.2602

60)/(05.360)/(22.1 sms

ssmssmvavg ≈×

×+×=

tvx 0=

Following the definition: (a) As the car moves in the same direction, so the total displacement is: The total time: So:

3. An automobile travels on a straight road for 40 km at 30 km/h. It then continues in the same direction for another 40 km at 60 km/h. (a) What is the average velocity of the car during this 80 km trip? (assume that it moves in the positive x direction) (b) What is the average speed? (c) Graph x versus t and indicate how the average velocity is found on the graph.

interval timentdisplaceme

=avgv

)(804040 kmx =+=Δ

)(26040

3040 ht =+=Δ

)/(40280 hkm

ΔtΔxvavg ===

(b) Following the definition: So: (c) We use the following equation to graph x versus t:

interval timedistance total

=avgs

)(804040 kmd =+=

)(26040

3040 ht =+=Δ

)/(40280 hkm

Δtdsavg ===

tvx 0=

Average speed = total distance/time interval

downup

downup

downdown

up

up

downup

downup

downupssss

sD

sD

DDttDD

s+

×=

+

+=

+

+= 2

)/(2.44 hkms =

downup DD =

4. A car travels up a hill at a constant speed of 35 km/h and returns down the hill at a constant speed of 60 km/h. Calculate the average speed for the round trip.

30. The brakes on your car can slow you at a rate of 5.2 m/s2. (a) If you are going 146 km/h and suddenly see a state trooper, what is the minimum time in which you can get your car under the 90 km/h speed limit? (The answer reveals the futility of braking to keep your high speed from being detected with a radar or laser gun). (b) Graph x versus t and v versus t for such a slowing. •  a=-5.2 m/s2; v0=146 km/h or v0=40.6 m/s; v1=90 km/h or v1=25 m/s (a) The minimum time tmin must match:

10 vatvv ≤+= tvv 01 ≤

−

aè

(s)0.32.5)6.4025(

avv

t 01min =

−

−=

−=

(b) x vs. t and v vs. t:

;2.6t40.6tat21tvx 22

0 −=+= 5.2t-40.6atvv 0 =+=

•  Visit Online Function Graphers, e.g.: http://www.function-grapher.com/index.php

(a < 0)

This is a freely falling object problem: We choose the positive direction is downward, the origin O at the roof y0 = 0, so: (a) When the stone hits the ground, we have y = 30 m: (discard t < 0) (b)

48. A hoodlum throws a stone vertically downward with an initial speed of 15.0 m/s from the roof of a building, 30.0 m above the ground. (a) How long does it take the stone to reach the ground? (b) What is the speed of the stone at impact?

200 2

1tvyy at++=

+ ground

O

y

0v!

28.92115t030 t++=

)(38.1 st =⇒

)/(5.2838.18.91500 smgtvatvv ≈×+=+=+=

50. At time t=0, apple 1 is dropped from a bridge onto a roadway beneath the bridge; somewhat later, apple 2 is thrown down from the same height. Figure below gives the vertical positions y of the apples versus t during the falling, until both apples have hit the roadway. With approximately what speed is apple2 thrown down?

• Apple 1 hits the roadway at t=2 s:

20 gt21yy −=

20 29.821y0 ××−= (m)6.19y0 =è

• Apple 2 is thrown down at t=1 s and hits the roadway at t=2.25 s:

200 gt

21tvyy −+= è

22200 gt

21tvy0 −+=

(s)1.25t2 = è (m/s)9.6v0 −≈

Chapter 1 Bases of Kinematics 1.1. Motion in One Dimension 1.2. Motion in Two Dimensions 1.2.1. The Position, Velocity, and Acceleration Vectors

1.2.2. Two-Dimensional Motion with Constant Acceleration.

Projectile Motion

1.2.3. Circular Motion. Tangential and Radial Acceleration

1.2.4. Relative Velocity and Relative Acceleration

Vectors (Recall) R1. Vectors and scalars: •  A vector has magnitude and direction; vectors follow certain

rules of combination. •  Some physical quantities that are vector quantities are

displacement, velocity, and acceleration. •  Some physical quantities that does not involve direction are

temperature, pressure, energy, mass, time. We call them scalars.

R2. Components of vectors: •  A component of a vector is the projection of the vector on an axis. •  If we know a vector in component notation (ax and ay), we determine it in magnitude-angle notation (a and θ):

ax=acosθ a y=asinθ

a= ax2 + ay

2 tanθ =ayax

R3. Adding vectors: R3.1. Adding vectors geometrically: •  Vector subtraction:

!s = !a +!b θ

s2 = a2 + b2 − 2abcos(180−θ )

!d = !a −

!b = !a + (−

!b)

R3.2. Adding vectors by components: R4. Multiplying a vector by a vector: R4.1. The scalar product (the dot product): R4.2. The vector product (the cross product): ϕ is the smaller of the two angles between and

!s = !a +!b

!a = axi + ay j;!b = bxi + by j

sx = ax + bx; sy = ay + by

!a•!b = abcosφ

!a•!b = axbx + ayby

!c = !a×!b

c = absinφ!a!b

The direction of is determined by using the right-hand rule: Your fingers (right-hand) sweep into through the smaller angle between them, your outstretched thumb points in the direction of In the right-handed xyz coordinate system:

!c !a!c

!b

!a = axi + ay j + azk!b = bxi + by j + bzk

!c = (aybz − byaz )i + (azbx − bzax ) j + (axby − bxay )k

!c = !a×!b

k = i × j

1.2. Motion in Two Dimensions 1.2.1. The Position, Velocity, and Acceleration Vectors Position and Displacement: y

x i

jr!

•  A particle is located by a position vector: jyixr +=

!

ix and jy are vector components of r!

•  Displacement: 12 rrr !!!−=Δ

)jyi(x)jyi(xrΔ 1122 +−+=!

jyixj)y-(yi)x-(xrΔ 1212 Δ+Δ=+=!

x and y are scalar components of r!

• Three dimensions: kzjyixr ++=!

kzjyixk)z-(zj)y-(yi)x-(xrΔ 121212 Δ+Δ+Δ=++=!

Average Velocity and Instantaneous Velocity:

intervaltimentdisplacemevelocityaverage =

trvavg Δ

Δ=!

!

•  Instantaneous Velocity, Δt→0:

dtrd

trlimv

0Δt

!!!

=ΔΔ

=→

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

jvivv

jdtdyi

dtdx)jyi(x

dtdv

yx +=

+=+=

!

!

The scalar components of v!

dtdyv,

dtdxv yx ==

Three dimensions:

kvjvivv zyx ++=!

dtdzvz =

Average Acceleration and Instantaneous Acceleration:

intervaltimeyin velocit changeonacceleratiaverage =

tvaavg Δ

Δ=!!

•  Instantaneous Acceleration, Δt→0:

dtvd

tvlima

0Δt

!!!=

ΔΔ

=→

a!The scalar components of jaiaa yx +=

!

,,dtdv

adtdva y

yx

x ==

Three dimensions:

dtdva z

z =

;kajaiaa zyx ++=!

1.2.2. Two-Dimensional Motion with Constant Acceleration. Projectile Motion Key point: To determine velocity and position, we need to determine x and y components of velocity and position

!a = axi + ay j = constant

!r = !r0 +!v0t +

12!at2 = xi+yj

!v = !v0 +!at = vx i + vy j

Along the x axis:

vx =v0x + axt; x = x0 + v0xt + 12

axt2

Along the y axis:

vy = v0y + a yt; y = y0 + v0yt + 12

ayt2

Sample Problem: A particle with velocity (m/s) at t=0 undergoes a constant acceleration of magnitude a = 3.0 m/s2 at an angle θ = 1300 from the positive direction of the x axis. What is the particle’s velocity at t=5.0 s, in unit-vector notation and in magnitude-angle notation?

j4.0i2.0v0 +−=!

a!

v!

•  Key issues: This is a two-dimensional motion, we must apply equations of straight-line motion separately for motion parallel

tavv;tavv y0yyx0xx +=+=v0x=-2.0 (m/s) and v0y=4.0 (m/s)

)(m/s2.30)sin(1303.0sinθaa

)(m/s-1.93)cos(1303.0cosθaa20

y

20x

=×==

=×==y

a!θ

•  At t= 5 s: (m/s) 15.5v;(m/s)-11.7v yx ==

j15.5i1.71v +−=!

• The magnitude and angle of : x (m/s)4.19vvv 2y

2x =+=

0

x

y 127θ1.33vv

)θtan( =⇒−≈=

v!

Projectile motion !a = constantax = 0; ay = −g

θ0: launch angle R: horizontal range

Projectile Motion: A particle moves in a vertical plane with some initial velocity but its acceleration is always the free-fall acceleration, the motion of this particle is called projectile motion.

•  Ox, horizontal motion (no acceleration, ax = 0):

x = x0 + v0xt = x0 + v0cosθ0t•  Oy, vertical motion (free fall, ay = -g if the positive y direction is upward):

2000 gt21-tsinθvyy +=

vy = v0y + ayt = v0sinθ0 -gt

• The equation of the path: ( )

( )200

2

0 θcosv2gx-xtanθy =

• Horizontal range: 0

20 sin2θgvR =

vx = v0x = v0cosθ0 = constant

Example: A projectile is shot from the edge of a cliff 115m above ground level with an initial speed of 65.0 m/s at an angle of 350 with the horizontal (see the figure below). Determine: (a) the maximum height of the projectile above the cliff; (b) the projectile velocity when it strikes the ground (point P); (c) point P from the base of the cliff (distance X).

x

y (a) At its maximum height: 0gt-sinθvv 00y ==

g00sinθvt =


( )max

200

2sinθvy Hg

==

Hmax

(m) 9.70max =H

Hmax x

(b) its velocity:

(m/s) 25.53cosθvv 00x ==

yaΔ=− 2vv 20y

2y

(m/s) 28.37sinθvv 000y ==

)(m/s 8.9 2−=a(m) 1150 −=−=Δ yyy P

(m/s) 36.60vy −=

j (m/s) 36.60i (m/s)25.53v −=!

(c) Calculate X: g

y00y

v-sinvtatsinvv

θθ =⇒+=

(s) 96.9=t(m) 37.530cosθvvX 00x === tt

Note: for (b), we can solve as follows:


0ytsinθvgt21

P002 =+−

0115t 28.734.9t2 =−−

quadratic equation:

(s) 96.9=⇒ tgtsinvatsinvv 00y −=+= θθ

(m/s) 33.60vy −=

then for (c), use t = 9.96 (s)

Sample Problem (page 70): Figure below shows a pirate ship 560 m from a fort defending the harbor entrance of an island. A defense cannon, located at sea level, fires balls at initial speed v0= 82 m/s. (a) At what angle θ0 from the horizontal must a ball be fired to hit the ship? (b) How far should the pirate ship be from the cannon if it is to be beyond the maximum range of the cannon balls?

0

20 sin2θgvR =

20

1

vgRsin2θ −=

00 63θor 27θ ≈≈

(a)

(b)

(m)686)452sin(8.9

82sin2θgvR

2

0

20

max =×==

20

22

10

11

000

cosθvRt;

cosθvRt

tcosθvxx

==⇒

+=

R1 R2

è It is likely answer 4

However

gθsin2vt;

gθsin2vt

t θsinvv

202

101

00y

==⇒

−= g when the shells hit the ships, 00y θsinvv −=

→θ1>θ2 è t2 < t1: the answer is B

the farther ship gets hit first

Vo

Vo θ1 θ2


Uniform Circular Motion: A particle moves around a circle or a circular arc at constant speed. The particle is accelerating with a centripetal acceleration:

Where r is the radius of the circle v the speed of the particle

rva2

=

vr2T π

= (T: period)


Tangential and radial acceleration: If the speed is not constant, then there is also a tangential acceleration.

tr aaa !!!+=

ra!

a!

ta!

Tangential acceleration Radial (centripetal) acceleration

The path of a particle’s motion

ra!

a!ta!

Homework: 6, 7, 11, 20, 27, 29, 54, 58, 66

(1) read sec. 2-10. (2) 1-6, 16, 20, 29-31, 33, 46, 48,...

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