info first exam thursday oct. 2 chapters 1, 2, and 3 homework due at that time. 3inch x 5 inch card...

InfoFirst Exam Thursday Oct. 2 Chapters 1, 2, and 3

homework due at that time.

3inch x 5 inch card is okay to use

1-side of equations, no sentences

1st lab due Monday Oct. 6

2nd lab Due Monday Oct. 13

.

Kinematic Description of Motion

•Speed

•Distance

•Position

•Displacement

•Velocity

•Acceleration

•Time

Kinematic Description of Motion

•Speed

•Distance

•Time

•all scalar quantities

Average speed = Distance traveled

Time taken

Avg. sp = d

t (m/s)

Average Speed

•Speed

•Distance

•Time

•all scalar quantities

Average speed = Distance traveled

Time taken

Avg. sp = d

t (m/s)

Instantaneous Speed

Instantaneous speed = Distance traveled

tiny time interval

sp = d

∆t (m/s)

Speed at an instant in time

How long does and instant last?

What is the average speed of a runner who runs 100.0 m in 11.0 sec?

•Given: d=100.0 m t=11.0 s

•Want: avg. sp

•Use: Avg. sp =

d

t

100.0 m

11.0 s = 9.09 m/s

A race car driver must average 100 mi/hr for a 200 mi trial? In the first

100 miles the driver only averages 50 mi/hr. With what average speed must she travel the last 100 miles in order

to qualify for the big race?

Displacement & Velocity

•Displacement

•velocity

•Both vectors (Magnitude and direction)

•Displacement: The straight line distance between two points, along with the direction from starting point to final point.

Displacement & Velocity

•Displacement (∆x) : The straight line distance between two points, along with the direction from starting point to final point.

•X1 position of point 1 , X2 position of point 2

x1

x2∆x=x2-x1

∆x

Displacement: Change in position

Displacement & Velocity

•Displacement

•velocity

•Both vectors

Average velocity = displacement

Time taken

V = ∆x

t (m/s), and direction

Displacement & Velocity•What is the average velocity of a car that travels from Vancouver to Seattle (160 miles north) in 4.00 hrs?

•Given ∆x=160 miles North, t=4.0 hr

•Want: V

•Use:

Displacement & Velocity•What is the average velocity of a car that travels from Vancouver to Seattle (160 miles north) in 4.00 hrs?

•Given ∆x=160 miles North, t=4.0 hr

•Want: V

•Use:

Average velocity = displacement

Time taken

V = ∆x

t = 40 mi/hr, N

160 miles, N

4.00 hr =

Displacement & Velocity•What is the average velocity of a car that travels from Vancouver to Seattle (160 miles north) in 4.00 hrs and back to Vancouver in 3.00 hrs?

•Given ∆x=0 miles North, t=7.0 hr

•Want: V

Displacement & Velocity•What is the average velocity of a car that travels from Vancouver to Seattle (160 miles north) in 4.00 hrs and back to Vancouver in 3.00 hrs?

•Given ∆x=0 miles North, t=7.0 hr

•Want: V

•Use:Average velocity =

displacement

Time taken

Displacement & Velocity•What is the average velocity of a car that travels from Vancouver to Seattle (160 miles north) in 4.00 hrs and back to Vancouver in 3.00 hrs?

•Given ∆x=0 miles North, t=7.0 hr

•Want: V

•Use:Average velocity =

displacement

Time taken

V = ∆x

t = 0 mi/hr, N

0 miles, N

7.00 hr =

Displacement & Velocity•What is the average speed of a car that travels from Vancouver to Seattle (160 miles north) in 4.00 hrs and back to Vancouver in 3.00 hrs?

•Given d=320 miles, t=7.0 hr

•Want: V

Displacement & Velocity•What is the average speed of a car that travels from Vancouver to Seattle (160 miles north) in 4.00 hrs and back to Vancouver in 3.00 hrs?

•Given d=320 miles, t=7.0 hr

•Want: V

•Use:Avg. sp =

distance

Time taken

Displacement & Velocity•What is the average speed of a car that travels from Vancouver to Seattle (160 miles north) in 4.00 hrs and back to Vancouver in 3.00 hrs?

•Given d=320 miles, t=7.0 hr

•Want: V

•Use:Avg. sp =

distance

Time taken

Avg. sp = d

t = 45.7 mi/hr

320 miles7.00 hr

=

Displacement & Velocity

v =x−xot

x=xo + vtx= vt

X position at time t

Xo initial position at t=0 (usually set to zero)

average velocityv

A car with a dripping gas tank travels along a road in a straight line. A drop falls from the gas tank every five seconds and over a section of road the spots from the gasoline can be used to determine the average speed of the car. From the spots in the figure below, the average speed of the car is determined to be a) 20 m/s, b) 24 m/s, c) 30 m/s, or d) 120 m/s?

QUICK QUIZ 2.1

(b). The average speed is the total distance divided by the total time elapsed. The distance between two successive spots represents a time interval of 5 seconds. There are therefore 5 time segments of 5 seconds so that the total time elapsed is 25 seconds. So vavg = x/t = 600 m/25 s = 24 m/s. It would be incorrect to simply count the number of spots and obtain a time equal to 6 X 5 = 30 seconds. Recall that it is the time interval, t, rather than the time, t, that belongs in the average velocity equation.

.

QUICK QUIZ 2.1 ANSWER

A car travels from Vancouver with an average speed of 50 mi/hr.

What can be said about where this car is after 1 hr?

A car travels from Vancouver with an average velocity of 50 mi/hr North for one hour .

What can be said about where this car is after 1 hr?

A car travels with an average speed of 50 mi/hr from Vancouver.What can be said about where this car is after 1 hr? It is somewhere wiithin or on the boundaries of a circle of radius 50 miles centered about vancouver.

A car travels from Vancouver with an average velocity of 50 mi/hr North for one hour .

What can be said about where this car is after 1 hr?It is 50 miles north of vancouver.

Average acceleration

Vxf velocity at time t

Vxi intial velocity at t=0.

€

a x =vxf −vxi

tor

vxf = vxi + a xt

€

a x

Average acceleration

What is the average acceleration of a jet that starts from rest and speeds up to 200 mi/hr in 8.0 sec?

€

a x

€

vxf = vxi + a xt

Average accelerationWhat is the average acceleration of a jet that starts from rest and speeds up to 200 mi/hr in 8.0 sec?

Given: Vo=0.0 (rest) V=200 mi/hr t=8.0 sWant: aUse:

€

vxf = vxi + a xt

€

a x

Average acceleration a Given: Vo=0.0 (rest) V=200 mi/hr t=8.0 sWant: a Use:

€

vxf = vxi + a xt

€

vxf = vxi + a xt

vxf − vxi = a xt

a x =vxf −vxi

t=

200mi / hr − 0

8.0s= 25mi / hr / s

Average acceleration a What is the average acceleration of a bird that starts from rest and speeds up to 20.0 m/s in 8.00 sec?

Given: Vxi=0.0 (rest), Vxf=20.0 m/s, t=8.0 s

Want: ax

Use:

€

vxf = vxi + a xt

Average acceleration a Given: Vo=0.0 (rest) V=20.0 m/s t=8.0 sWant: a Use:

€

vxf = vxi + a xt

€

vxf = vxi + a xt

vxf − vxi = a xt

a x =vxf −vxi

t=

20.0m / s −0

8.0s= 2.50m / s / s = 2.5

m

s2

Is it possible to accelerate while traveling at a constant speed?

A dragster starts from rest and takes off down a race track. At the precise moment that the dragster starts to move, its initial velocity is zero. At this precise moment, the instantaneous acceleration of the dragster is a) in the same direction as the subsequent motion of the dragster, b) in the opposite direction to the subsequent motion of the dragster, or c) zero?

QUICK QUIZ 2.2

(a). The instantaneous acceleration is the derivative of the velocity with respect to time. Examining a possible graph of velocity versus time for the dragster reveals that, although the velocity is zero at t = 0, the derivative of the velocity (or slope of the graph) is not zero and is positive, indicating that the acceleration is non-zero and in the direction of the motion.

QUICK QUIZ 2.2 ANSWER

Velocity vs time graph(acceleration is the slope )

Velocity vs time graph(acceleration is the slope )

Velocity vs time graph(acceleration is the slope )

Fig. 2.5, p.31

When a car skids after applying its brakes, the acceleration of the car is in the opposite direction to the velocity of the car. If the acceleration of the car would remain constant in this direction, the car would a) eventually stop and remain stopped, (b) eventually stop and then start to speed up in the forward direction, (c) eventually stop and then start to speed up in the reverse direction, or (d) never stop but continue to speed up in the forward direction?

(end of section 2.4)QUICK QUIZ 2.3

(c). If the acceleration would remain constant in a direction opposite the initial velocity direction, eventually the car would stop and then speed up in the reverse direction. A way to achieve such a feat would be to slam the car into reverse while traveling forward. The car would at first skid, then stop, and then speed up in reverse. For a constant reverse acceleration, the velocity curve is graphed below where it can be seen that the car instantaneously stops when the velocity is zero and then continues with a negative velocity.

QUICK QUIZ 2.3 ANSWER

A car traveling with a velocity of +30.0 m/s accelerates at a rate of -4.0 m/s2. How long will it take this car to stop?

Given vo = +30.0 m/s, a=-4.0 m/s2 and v = 0.0

Want: t

Use:

€

vxf = vxi + a xt

More about units

m / sm / s2 =

mss2

m=s

Kinematic equations

1

2

3

4 Constant a

5

€

x f = xo + v xt

vxf = vxi + a xt

v x =vxf + vxi

2

x f = xo + vxit +12

axt2

vxf2 = vxi

2 + 2ax (x f − xo )

Distance=Avg Sp (time)

V vs t Graph and average Velocity

t

Vo

Constant Velocity ax=__?___

v =Vxi

V

X=Vxit = “AREA”

V vs t Graph constant acceleration

V

t

Vo

A straight line

€

X = ΔX1 + ΔX2 + ΔX3 +• • • = ΔXi∑ = V iΔt∑ =" AREA"

€

x f = xo + v xt

Δx = v xt

€

v x =ΔX

t=

" AREA"t

V vs t Graph Constant a

t

€

v x =vxf + vxi

2

€

vxf

€

vxi

€

v x =ΔX

t=

" AREA"t

=

vxf + vxi

2 ⎛ ⎝ ⎜

⎞ ⎠ ⎟t

t=

vxf + vxi

2 ⎛ ⎝ ⎜

⎞ ⎠ ⎟

1

2

3

€

x f = xo + v xt

vxf = vxi + a xt

v x =vxf + vxi

2Constant a

Eliminate v from 1,2, &3

(use 3)

(use 2)

algebra

4

€

x = v t

Δx =vxf + vxi

2

⎛ ⎝ ⎜

⎞ ⎠ ⎟t

Δx =vxi + a xt + vxi

2 ⎛ ⎝

⎞ ⎠t

Δx =2vxi + a xt

2

⎛

⎝ ⎜

⎞

⎠ ⎟t

Δx = vxit +12

a xt2

x f = xi + vxit +12

a xt2

Eliminate t from 1,2, &3

(use 3)

(use 2)

algebra

5

€

x = v t

Δx =vxf + vxi

2 ⎛ ⎝ ⎜

⎞ ⎠ ⎟t

Δx =vxf + vxi

2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

vxf −vxi

ax

⎛ ⎝ ⎜

⎞ ⎠ ⎟

Δx =vxf

2 −vxf vxi + vxf vxi − vxi2

2ax

⎛

⎝ ⎜

⎞

⎠ ⎟

2axΔx = vxf2 − vxi

2

vxf2 = vxi

2 + 2axΔx

Kinematic equations

1

2

3

4 Constant a

5

€

x f = xo + v xt

vxf = vxi + a xt

v x =vxf + vxi

2

x f = xo + vxit +12

axt2

vxf2 = vxi

2 + 2ax (x f − xo )

Distance=Avg Sp (time)

Differential equations

€

vx =dx(t)

dt

ax =dv(t)

dt

x = vxdt∫

v = axdt∫

A jet aircraft being launched from an aircraft carrier is accelerated from rest along a 94-meter track for 2.5 s. What is the launch speed of the aircraft?

A rock hits the ground at a speed of 10 m/s and leaves a hole 25 cm deep. What was the acceleration of the rock while coming to a stop.

1

2

3

4 Constant a

5

€

x f = xo + v xt

vxf = vxi + a xt

v x =vxf + vxi

2

x f = xo + vxit +12

axt2

vxf2 = vxi

2 + 2ax (x f − xo )

Distance=Avg Sp (time)

Free Fall a=9.8m/s2 (downward)

Choose +x direction to be in the direction of initial motion

3. The position versus time for a certain particle moving along the x axis isshown in Figure P2.3. Find the average velocity in the time intervals (a) 0 to 2 s,(b) 0 to 4 s, (c) 2 s to 4 s, (d) 4 s to 7 s, (e) 0 to 8 s.

4. A particle moves according to the equation x = 10t2 where x is in metersand t is in seconds. (a) Find the average velocity for the time interval from 2.00 sto 3.00 s. (b) Find the average velocity for the time interval from 2.00 to 2.10 s.

9. Find the instantaneous velocity of the particle described in Figure P2.3 atthe following times: (a) t = 1.0 s, (b) t = 3.0 s, (c) t = 4.5 s, and (d) t = 7.5 s.

12. A particle starts from rest and accelerates as shown in Figure P2.12. Determine (a) the particle's speed at t = 10.0 s and at t = 20.0 s, and (b) the distance traveled in the first 20.0 s.

15. A particle moves along the x axis according to the equationx = 2.00 + 3.00 t – 1.00 t2, where x is in meters and t is in seconds. At t = 3.00 s,find (a) the position of the particle, (b) its velocity, and (c) its acceleration.

29. The driver of a car slams on the brakes when he sees a tree blocking theroad. The car slows uniformly with an acceleration of –5.60 m/ s2 for 4.20 s,making straight skid marks 62.4 m long ending at the tree. With what speeddoes the car then strike the tree?

30. Help! One of our equations is missing! We describe constant-accelerationmotion with the variables and parameters vxi, vxf, ax, t, and xf – xi. Of theequations in Table 2.2, the first does not involve xf – xi . The second does notcontain ax; the third omits vxf and the last leaves out t. So to complete the setthere should be an equation not involving vxi . Derive it from the others. Use it tosolve problem 29 in one step.

33. An electron in a cathode ray tube (CRT) accelerates from 2.00 × 104 / m s to6.00 × 106 / m s over 1.50 c . (m a) Ho w lon g doe s the electron take to travel this1.50 cm? (b) What is its accelerati ?on

38. Speedy Sue, driving at 30.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 155 m ahead traveling at 5.00 m/s. Sue applies her brakes but can accelerate only at 2.00 m/s2 because the road is wet. Will there be a collision? If yes, determine how far into the tunnel and at what time the collision occurs. If no, determine the distance of closest approach between Sue's car and the van.

43. A student throws a set of keys vertically upward to her sorority sister, who is in a window 4.00 m above. The keys are caught 1.50 s later by the sister's outstretched hand. (a) With what initial velocity were the keys thrown? (b) What was the velocity of the keys just before they were caught?

49. A daring ranch hand sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed of the horse is 10.0 m/s, and the distance from the limb to the saddle is 3.00 m. (a) What must be the horizontal distance between the saddle and limb when the ranch hand makes his move? (b) How long is he in the air?

52. A freely falling object requires 1.50 s to travel the last 30.0 m before it hits the ground. From what height above the ground did it fall?

70. A rock is dropped from rest into a well. (a) The sound of the splash is heard 2.40 s after the rock is released from rest. How far below the top of the well is the surface of the water? The speed of sound in air (at the ambient temperature) is 336 m/s. (b) What If? If the travel time for the sound is neglected, what percentage error is introduced when the depth of the well is calculated?