A train moves along a straight track and its position vs. time … · · 2014-10-20Distance...

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A train moves along a straight track and its position vs. time looks like:

Which graph best depicts its velocity vs. time?

a

b c

d

e

a

b

c

d

e

Slope of the tangent line

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Assignments, etc. Note: •  My class notes (what’s projected on the screen) are placed on the

course web site the evening before class and then replaced with a version containing answers after class sometime.

•  Bring the blue tutorial book to recitation this week. Hand in your HW at the beginning of recitation.

•  BOOKS? Pdfs placed on D2L under “Tutorial Scans”.

You should have already: •  Read Chs. 1 & 2 of Wolfson’s text & Prof. Dubson’s notes.

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For this week: •  Begin reading Ch. 3. •  You’ve done CAPA set #1; begin CAPA set #2. •  Complete this week’s recitation homework.

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Kinematics in 1D (Ch. 2)

x x x

t t t

Constant velocity: Slowing down Speeding up + positive - negative 0 zero (standing still)

(+)

(-)

What if v ≠ constant?

5 L4 W 9/3/14


v = limΔt→0

ΔxΔt

= dxdt

≡ instantaneous velocity = slope of the tangent line

Note: instantaneous speed = s ≡ v Does s = v ?

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Need a coordinate system: origin, positive direction in 1D. Position = x Distance traveled = d Displacement = Δx = xf – xi Average velocity = Instantaneous velocity = Instantaneous speed = | v | Graphical relation between position vs. time and velocity vs. time: Velocity is the local slope of the tangent line on a x vs t graph.

v = ΔxΔt

v = limΔt→0

ΔxΔt

= dxdt

Average speed = distance traveledΔt


What we’ve seen so far:

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The graph shows positions versus time for two trains running on parallel tracks. Which statement is true?

A) At time tC, both trains have the same velocity. B)  Both trains speed up all the time. C) Both trains have the same velocity at some time before tC. D) At some time, both trains have the same acceleration. E)  None of the above statements is true.

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If velocity is changing, there is “acceleration”. Acceleration is the time rate of change of velocity

If Δt is large: a = average acceleration = If Δt is very small: a = instantaneous acceleration:

a

a = limΔt→0

ΔvΔt

= dvdt


Units: m / ss

= ms2a = change in velocity

elapsed time= ΔvΔt

=vf − vit f − ti

= v2 − v1

t2 − t1

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x(t)

v(t) = dxdt

a(t) = dvdt

= d2xdt 2

position versus time for an object velocity is the time derivative of x(t) acceleration is the time derivative of v(t) and the second

derivative of x(t)

Kinematics in 1D (Ch. 2) A little math:

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x(t)

v(t) = dxdt

a(t) = dvdt

= d2xdt 2


derivative of x(t)

x

t

slowing down

1

2

dvdt

< 0 or d2xdt 2 < 0


An object is slowing down continuously with the position versus time plot shown at left. It’s acceleration a is A) > 0 B) < 0 C) = 0.

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x(t)

v(t) = dxdt

a(t) = dvdt

= d2xdt 2


derivative of x(t)

x

t

x

t

slowing down speeding up

1

2

dvdt

< 0 or d2xdt 2 < 0 dv

dt> 0 or d

2xdt 2 > 0


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x(t)

v(t) = dxdt

a(t) = dvdt

= d2xdt 2

Derivatives you might be expected to know: ddtb = 0 (b = const) d

dtt n = ntn−1

ddt

sin(bt) = bcos(bt) ddt

cos(bt) = −bsin(bt)ddt

(bt 2 + ct + d) = 2bt + c (b,c,d = const)


derivative of x(t)


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x(t)

v(t) = dxdt

a(t) = dvdt

= d2xdt 2

Derivatives you might be expected to know:


derivative of x(t)

ddtb = 0 (b = const) d

dtt n = ntn−1

ddt

sin(bt) = bcos(bt) ddt

cos(bt) = −bsin(bt)


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ddtt n = ntn−1A car accelerates on a straight track and its position

changes with time quadratically according to:

x(t) = 5t 2

x

t

What will its graph of velocity versus time look like? A) A straight line with a positive slope = 10. B) A straight line with a negative slope = -10. C) A straight line with a positive slope = 5. D) A straight line with a negative slope = -5. E) A straight line with a slope = 0.

v(t) = 10t

t

v

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x(t) = 5t 2

x

t

What will its graph of acceleration versus time look like? A) A straight line with a positive slope = 10. B) A straight line with a negative slope = -10. C) A straight line with a positive slope = 5. D) A straight line with a negative slope = -5. E) A straight line with a slope = 0.

v(t) = 10t

t

v

t

a a(t) = 10

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x(t) = 5t 2

x

t

v(t) = 10t

t

v

t

a a(t) = 10

Therefore, for an object that moves with a constant acceleration: its velocity changes linearly with time, its position changes quadratically (or parabolically) with time.

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Examples of constant acceleration motion: (1) Objects that are speeding up or slowing down in a uniform way. (2) Glider on an airtrack. (3) Object in “free fall” near the surface of the earth. Free fall = the only force acting on the body is gravity.

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At time t1= 0 s, a car is moving with velocity -15 m/s (i.e., to the left). At later time t2=5 s, the car is moving with velocity – 5 m/s.

What is the direction of acceleration?

A)  Right and positive. B)  Left and negative. C)  There is no acceleration. D)  Right and negative. E)  Left and positive.

a = ΔvΔt

=v2 − v1t2 − t1

=(−5 − (−15))m / s

(5 − 0)s= +2 m

s2

Note: the direction of acceleration is not the direction of velocity.

+x

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At time t1= 0 s, a car is moving with velocity -15 m/s (i.e., to the left). At later time t2=5 s, the car is moving with velocity – 5 m/s.

5

t

v

-5 -10 -15


(m/s)

(s) 5

The slope (acceleration) here is positive though the velocity is negative.

+x

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Object moves in the +x direction v > 0 ⇒Object moves in the -x direction v < 0 ⇒

Why? Because v ≈ ΔxΔt

What determines the sign of acceleration? Direction of motion? NO!

The change in velocity: a ≈ Δv

Δt

Note: change in velocity may not be in the same direction as motion

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1 2 3

4 5

The arrow’s length is proportional to the instantaneous speed and its direction points in the direction of motion. Vector: has a magnitude & direction

v

Up

Back

Look at motion up and back a tilted air track:

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At the moment when velocity is zero, what is the sign of acceleration? A) Positive B) Negative C) Zero

1

2

3

4

5

+x

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1

2

3

4

5

a = dvdt

< 0 (negative)


+x

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2 4

Just before the top Just after the top

Another way to look at this – vector analysis:

v1

v2

What do you have to add to to get ? v1

v2

Δv = v2 −v1

∴ v2 =v1 + Δv

+x

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Another way to look at this – vector analysis:

v1

v2

What do you have to add to to get ? v1

v2

Δv

v2 =v1 + Δv

+x

A train moves along a straight track and its position vs. time … · · 2014-10-20Distance...

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