Chapter 2Kinematics in one dimension

Galileo - the first modern kinematics

1) In a medium totally devoid of resistance all bodies will fall at the same speed

2) During equal intervals of time, a falling body receives equal increments of speed

1564 - 1642

I greatly doubt that Aristotle ever tested by experiment whether it be true that two stones, one weighing 10 times the other, if allowed to fall at the same instant from a height of 100 cubits, would so differ in speed that when the heavier had reached the ground the other would not have fallen more than 10 cubits. !

I, who have made the test can assure you that a cannon ball weighing one or two hundred pounds, or more, will not reach the ground by as much as a span ahead of a musket ball weighing only half a pound. !

Salviati in Two New Sciences, 1636, by Galileo

Idealization and extrapolation !

- Totally devoid of resistance was not possible (nature abhors a vacuum), so Galileo made things worse: !

- If deviations from his law are far worse in a dense medium than a thin one, is it not reasonable to suppose that they would disappear altogether if the medium were absent?

Describing motion• Complete description gives position as a

function of time: x(t) – e.g. x=(10m/s) t

• If t = 1 s, x = 10 m • If t = 2 s, x = 20 m !

– e.g. x = a t2 =(10 m/s2) t2

• If t = 1 s, x = 10 m • If t = 2 s, x = 40 m

Describing motion• Complete description gives position as a

function of time: x(t) • But speed and acceleration interesting too:

– Want v(t), a(t) – Also v(x) – Should be derivable from x(t)

1) Displacement• Vector from initial to final position

x0 + Δx = xΔx = x − x0

• In 1d, the sign indicates direction:

Δx > 0, displacement to rightΔx < 0, displacement to leftΔx = 0, no displacement

2) Speed and Velocity• Time rate of change of position • Average speed = distance/time = s/∆t (scalar)

• Average velocity = displacement/time (vector)

v = ΔxΔt

Units for speed or velocity: m/s

t

x

∆x

∆t

graphical representation x vs t

• Constant velocity: equal intervals of distance in equal intervals of time --> slope of a straight line

v = ΔxΔt

• Constant velocity: displacement is area under v vs t curve

= area

t

v

graphical representation v vs t

Δt

v

Δx = vΔt

• Changing velocity: average speed depends on time interval -- curve on x vs t graph

• Here velocity increases in time

t

x

• Instantaneous velocity

v = limΔt→0

ΔxΔt

t

xRepresents tangent to x vs t curve.

• For an arbitrary v(t), in a small interval, the displacement is the area (approx.):

t

v

Δxi = viΔti

Δti

vi

• so the total displacement is the total area (if the intervals are small enough):

t

v

Δx = Δx1 + Δx2 + ...

• In particular, if v vs t is a straight line (say, v = at): – displacement is area of a triangle (here ∆x = x - 0)

t

v

x = 12(base)(height) = 1

2at 2

at

3) Acceleration• Time rate of change of velocity • Average acceleration

!a = Δ!vΔt

=!v − !v0t − t0

• Units: m/s2

• Constant acceleration: slope of a straight line (v vs t)

t

x

∆x

∆t

v

Δv a = ΔvΔt

• For an object starting from rest, v = at

• Changing acceleration: average acceleration depends on time interval -- curve on v vs t graph

• Here acceleration increases in time

t

xv

• Instantaneous acceleration

Represents tangent to v vs t curve.

t

xv

a = limΔt→0

ΔvΔt

• Positive acceleration: – acceleration in pos. dir’n – speeding up if v > 0

– slowing down if v < 0 – (deceleration)

v

t

t

v

• Negative acceleration: – acceleration in neg. dir’n – slowing down if v > 0 – (deceleration)

– speeding up if v < 0 – (acceleration)

t

v

v

t

4) Kinematic equations (constant acceleration)

• Quantities: x, v, a, t (and their initial values, x0, v0, a0, t0) – For constant acceleration, a = a0

– Usually choose coord system so x = x0 = 0 at t = t0 = 0 – Leaves 5 quantities typically (v0 not zero in general) – Want x(t), v(t), which give v(x), x(v,t)

• Acceleration: a = ∆v/∆t = constant !

• Velocity:

a = v − v0t − t0

= v − v0t

v = v0 + at

v

t

v0

• Position (displacement) – evaluate area

v

t

v0Δx = v0t +12(v − v0 )t

Δx = v0t + 12 at

2

x = x0 + v0t + 12 at

2

– Taking x0 = 0,

x = v0t + 12 at

2

v = v0 + at 1 all but x

x = v0t + 12 at

2 2 all but v

Eliminate t from 1 and 2:

3v2 = v02 + 2ax all but t

Eliminate a from 1 and 2: x = 1

2 (v + v0 )t = vt 4 all but a

x = vt − 12 at

2 5 all but v0Eliminate v0 from 1 and 2:

5) Applications• Draw a picture

– use it to identify kinematic quantities

• Choose +ve direction and origin – (for each object)

• Identify quantities that are given and that are needed

• Identify the equations that relate them • Solve

2.28 (a) What is the magnitude of the average acceleration of a skier who, starting from rest, reaches a speed of 8.0 m/s when going down a slope for 5.0 s?   (b) How far does the skier travel in this time? !!A person is 11 m from a bus as it starts to leave the stop with an acceleration of 1.0 m/s2. If the person runs after the bus at 5.0 m/s, how long does it take to catch the bus (assuming the driver has not taken Newhart’s driving course)? !

Examples

Homework• C&J 2.35 (hint: when the two knights collide, they are at

the same position at the same time. One knight has positive acceleration and one has negative acceleration

• First graded WileyPlus assigned today, due a week Monday

C&J 2.35 In a historical movie, two knights on horseback start from rest 88.0 m apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 0.300 m/s2, while Sir Alfred's has a magnitude of 0.200 m/s2. Relative to Sir George's starting point, where do the knights collide?

5) Free-fall• Acceleration toward earth independent of mass,

and (nearly) constant (neglecting air resistance) !

• acceleration due to gravity: !

• g = 9.8 m/s2

A stone is dropped from the top of a tall building. After 3.00s of free fall, what is the displacement y of the stone?

y a v v t? -9.80 m/s 0 m/s 3.00 s

y = v0t + 12 at

2

= -44.1 m

• Maximum height

y a v v t? -g 0 m/s v

ymax =v02

2g

• Symmetry What is the speed at the original position?

v2 − v02 = 2ay v2 = v0

2 v = ±v0If y = 0, then

y = v0t − 12 gt

2

v = v0 − gt

a = −g

y

t

C&J problem: A woman on a bridge 75.0 m high sees a raft floating at constant speed on the river below. Trying to hit the raft, she drops a stone from rest when the raft has 7.00 m more to travel before passing under the bride. The stone hits the water 4.00 m in front of the raft. Find the speed of the raft.

C&J problem 2-45 The drawing shows a device that you can make with a piece of cardboard, which can be used to measure a person’s reaction time. Hold the card and suddenly drop it. Ask a friend to try and catch the card between his or her thumb and index finger. Initially, your friend’s fingers must be level with the asterisks at the bottom. By noting where your friend catches the card, you can determine his or her reaction time in milliseconds (ms). Calculate the distances d1, d2, and d3.

Homework• Problem 2.47 • Tutorial test this week • first Wiley test up, due Monday 11 pm

C&J 2-47 Review Conceptual Example 15 before attempting this problem. Two identical pellet guns are fired simultaneously from the edge of a cliff. These guns impart an initial speed of 30 m/s to each pellet. Gun A is fired straight upward, with the pellet going up and then falling back down, eventually hitting the ground beneath the cliff. Gun B is fired straight downward. In the absence of air resistance, how long after pellet B hits the ground does pellet A hit the ground?