vectors in mechanics
DESCRIPTION
Newtonian mechanics in vector form: Elements of vectors, operations on vectors, kinematics, Laws of Newton, Momentum & ImpulseTRANSCRIPT
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Vectors
in mechanics
Photo: Wikimedia.org
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Quantities
Vectors in mechanics 2
Scalar quantities Vector quantities
mass m (kg)
energy E (J)
power P (W)
displacement 𝑠 (𝑚)
velocity 𝑣 (𝑚𝑠−1)
acceleration 𝑎 (𝑚𝑠−2)
force 𝐹 (𝑁)
momentum 𝑝 (𝑘𝑔𝑚𝑠−1) torque 𝑇 (𝑁𝑚)
• Magnitude • Magnitude• Direction• Point of application
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Vectors in mechanics 3
Vector notation
𝐹 arrow above orF bold face
6.0 𝑐𝑚
30 𝑁
6.0𝑐𝑚 ≜ 30𝑁
1.0𝑐𝑚 ≜ 5.0𝑁 Scale
Head
Tail
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Point of application matters,…
Vectors in mechanics 4
𝐹
𝐹
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but…is often compromised…
Vectors in mechanics 5
𝐹𝑔
𝐹𝑎𝑖𝑟
𝐹𝑟𝑜𝑎𝑑
𝐹𝑛
𝐹𝑔
𝐹𝑟𝑜𝑎𝑑
𝐹𝑛
𝐹𝑎𝑖𝑟
𝐹𝑔
𝐹𝑟𝑜𝑎𝑑
𝐹𝑛
𝐹𝑎𝑖𝑟
…to make life easier
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Basic operations with vectors
Vectors in mechanics 6
Adding vectors 𝐹𝑛𝑒𝑤 = 𝐹𝑜𝑙𝑑1 + 𝐹𝑜𝑙𝑑2
Subtracting vectors 𝐹𝑛𝑒𝑤 = 𝐹𝑜𝑙𝑑1 − 𝐹𝑜𝑙𝑑2
Multiplying a vector with a scalar 𝐹𝑛𝑒𝑤 = 𝑎 ∙ 𝐹𝑜𝑙𝑑
Resolving a vector into components 𝐹𝑛𝑒𝑤1 + 𝐹𝑛𝑒𝑤2 = 𝐹𝑜𝑙𝑑
a = 2
𝜑𝐹 ∙ 𝐶𝑜𝑠 𝜑
𝐹 ∙ 𝑆𝑖𝑛 𝜑 𝐹
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Resolve a vector along 2 working lines
Vectors in mechanics 7
𝐹𝑛𝑒𝑤1
𝐹𝑛𝑒𝑤2
𝐹𝑜𝑙𝑑
copy
copy
𝐹𝑜𝑙𝑑 = 𝐹𝑛𝑒𝑤1+ 𝐹𝑛𝑒𝑤2
𝐹𝑛𝑒𝑤1 =?
𝐹𝑛𝑒𝑤2 =?
𝐹𝑜𝑙𝑑
2 given working lines
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Displacement vs. Distance
Vectors in mechanics 8
Type equation here.
𝑠1
𝑠2 = 𝑠1 + ∆ 𝑠
∆ 𝑠 = 𝑠2 − 𝑠1
vector scalar
Do not interpret ‘s’as the distance, because it represents the magnitude of the displacement vector 𝑠
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Velocity vs. Speed
Vectors in mechanics 9
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
𝑡𝑖𝑚𝑒𝑠𝑝𝑒𝑒𝑑 =
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑡𝑖𝑚𝑒
vector scalar
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Average vs. Instantaneous
Vectors in mechanics 10
t(s)
s(m)
∆𝑠
∆𝑡
𝑣𝑎𝑣𝑔 5𝑠 → 10𝑠 =𝑠 10 − 𝑠 5
10 − 5=∆𝑠
∆𝑡
= gradient of the secant line= difference quotient
= gradient of the tangent= differential quotient
𝑣 5𝑠 = lim𝑑𝑡→0
𝑠 5 + 𝑑𝑡 − 𝑠(5)
5 + 𝑑𝑡 − 5=𝑑𝑠
𝑑𝑡= 𝑠′
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Acceleration – notation issues
Vectors in mechanics 11
𝑎 =∆𝑣
∆𝑡=𝑣 − 𝑢
∆𝑡
𝑎 =∆𝑣
∆𝑡=𝑣2 − 𝑣1∆𝑡
Initial velocity = 𝑢
Initial velocity = 𝑣0
Movement along a straight line:• 𝑠 → 𝑠 𝑜𝑟 𝑥• 𝑣 → 𝑣• 𝑎 → 𝑎• Choose an origin• Choose a + direction!
𝑣
𝑠 +
me
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𝑎 =∆𝑣
∆𝑡⇒ ∆𝑣 = 𝑎 ∙ ∆𝑡 ⇒ 𝑣 − 𝑣0 = 𝑎 ∙ 𝑡 − 𝑡0 ⇒ 𝑣 = 𝑣0 + 𝑎 𝑡 − 𝑡0
Uniformly accelerated motion - I
Vectors in mechanics 12
𝑡
𝑣
𝑡𝑜
𝑣𝑜
𝑎 > 0
𝑡
𝑣
𝑡𝑜
𝑣𝑜
𝑎 < 0
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Uniformly accelerated motion - II
Vectors in mechanics 13
𝑡
𝑣
𝑡1
𝑣1
𝑡2
𝑣2
𝑣𝑎𝑣𝑔
𝑡
𝑣
𝑡1
𝑣1
𝑡2
𝑣2
𝑣𝑎𝑣𝑔
𝑣𝑎𝑣𝑔 =𝑣1 + 𝑣22
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𝑣𝑎𝑣𝑔 =∆𝑠
∆𝑡⇒ ∆𝑠 = 𝑣𝑎𝑣𝑔 ∙ ∆𝑡
Uniformly accelerated motion - III
Vectors in mechanics 14
𝑣𝑎𝑣𝑔 =𝑣 + 𝑣02
∆𝑠 =𝑣 + 𝑣02
∙ ∆𝑡
𝑣 = 𝑣0 + 𝑎 ∙ ∆𝑡
∆𝑠 =𝑣0 + 𝑎 ∙ ∆𝑡 + 𝑣0
2∙ ∆𝑡 = 𝑣0 + 1 2𝑎 ∙ ∆𝑡 ∙ ∆𝑡 ⇒ 𝑠 = 𝑠0 + 𝑣0 ∙ 𝑡 − 𝑡0 + 1 2 𝑎 ∙ 𝑡 − 𝑡0
2
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Uniformly accelerated motion - IV
Vectors in mechanics 15
𝑠 = 𝑠0 + 𝑣0 ∙ 𝑡 − 𝑡0 + 1 2𝑎 ∙ 𝑡 − 𝑡02
𝑡
𝑣
𝑡𝑜
𝑣𝑜
𝑎 > 0
𝑣 = 𝑣0 + 𝑎 𝑡 − 𝑡0
𝑡
𝑠
𝑡𝑜
𝑠𝑜
Gradient of tangent = 𝑣0
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Uniformly accelerated motion - V
Vectors in mechanics 16
𝑠 = 𝑠0 + 𝑣0 ∙ 𝑡 − 𝑡0 + 1 2𝑎 ∙ 𝑡 − 𝑡02
𝑡
𝑣
𝑡𝑜
𝑣𝑜
𝑎 < 0
𝑣 = 𝑣0 + 𝑎 𝑡 − 𝑡0
𝑡
𝑠
𝑡𝑜
𝑠𝑜
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Uniformly accelerated motion - VI
Vectors in mechanics 17
𝑠 = 𝑠0 + 𝑣0 ∙ 𝑡 − 𝑡0 + 1 2𝑎 ∙ 𝑡 − 𝑡02
𝑣 = 𝑣0 + 𝑎 𝑡 − 𝑡0General equations
𝑡0 = 0 Generally
𝑠0 = 0 Accelerationfrom rest𝑣0 = 0
𝑠 = 𝑠0 + 𝑣0 ∙ 𝑡 + 1 2𝑎 ∙ 𝑡2
𝑣 = 𝑣0 + 𝑎𝑡
𝑠 = 1 2𝑎 ∙ 𝑡2
𝑣 = 𝑎𝑡
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The math behind kinematics
Vectors in mechanics 18
𝑎 𝑡 =𝑑𝑣
𝑑𝑡= 𝑣′ 𝑡 = 𝑠′′ 𝑡
𝑣 𝑡 =𝑑𝑠
𝑑𝑡= 𝑠′ 𝑡
𝑠0
𝑠
𝑑𝑠 = 𝑡0
𝑡
𝑣 ∙ 𝑑𝑡
𝑣0
𝑣
𝑑𝑣 = 𝑡0
𝑡
𝑎 ∙ 𝑑𝑡
𝑠 𝑡 Displacement
Velocity
Acceleration 𝑎 𝑡
Differentiation
Differentiation
Integration
Integration
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Displacement diagrams
Vectors in mechanics 19
𝑡
𝑠
𝑡
𝑠
∆𝑡
∆𝑡
∆𝑠
∆𝑠
Determine 𝑡 read 𝑡-axis
Determine 𝑠 read 𝑠-axis
Determine 𝑣 gradient of the tangent
Determine 𝑎
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Velocity diagrams
Vectors in mechanics 20
Determine 𝑡 read 𝑡-axis
Determine 𝑠 area under the graph
Determine 𝑣 read 𝑣-axis
Determine 𝑎 gradient of the graph
𝑡
𝑣1
𝑡
𝑣
𝑡1 𝑡2
𝑡1 𝑡2
𝑣2
∆𝑣
∆𝑡
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Acceleration diagrams
Vectors in mechanics 21
𝑎
𝑎
Determine 𝑡 read 𝑡-axis
Determine 𝑠
Determine 𝑣 area under the graph
Determine 𝑎 read 𝑎-axis
𝑡1 𝑡2
𝑡1 𝑡2
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Relative velocity - I
Vectors in mechanics 22
How much time does it take A to catch up on B?
𝑣𝐴𝐵 = 𝑣𝐴 − 𝑣𝐵 = 15 − 10 = 5𝑚𝑠−1 ⇒ ∆𝑡 = ∆𝑠𝐴𝐵
𝑣𝐴𝐵 = 1005 = 20 𝑠
𝑣𝐴 = 15 𝑚𝑠−1 𝑣𝐵 = 10 𝑚𝑠
−1
100 𝑚
How much time does it take A to ‘meet’ B?
𝑣𝐴𝐵 = 𝑣𝐴 − 𝑣𝐵 = 15 − −10 = 25𝑚𝑠−1 ⇒ ∆𝑡 = ∆𝑠𝐴𝐵𝑣𝐴𝐵 = 150
25 = 6.0 𝑠
𝑣𝐴 = 15 𝑚𝑠−1 𝑣𝐵 = 10 𝑚𝑠
−1
150 𝑚
A
A
B
B
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Relative velocity - II
Vectors in mechanics 23
How fast does A approach B? 𝑣𝐴 = 15 𝑚𝑠−1
𝑣𝐵 = 10 𝑚𝑠−1
B
30°
30°
𝑣𝐵
𝑣𝐴
𝑣𝐴𝐵
𝑣𝐴𝐵𝑥
𝑣𝐴𝐵𝑦
𝑣𝐴𝐵𝑥 = 15𝐶𝑜𝑠 30° − 10 = 3.0 m𝑠−1
𝑣𝐴𝐵𝑦 = 15𝑆𝑖𝑛 30° = 7.5 m𝑠−1
𝑣𝐴𝐵 = 3.02 + 7.52 = 8.1 𝑚𝑠−1
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Forces
Vectors in mechanics 24
Name Symbol Value Direction Point of application
Weight 𝐹𝑊 𝐹𝑊 = m ∙ 𝑔 To the centre of the Earth Center of mass
Friction 𝐹𝐹 𝐹𝐹 = 𝑓 ∙ 𝐹𝑁 ∥ to surface Contact surface
Air resistance 𝐹𝐴𝑖𝑟 𝐹𝐴 = 1 2𝜌𝐶𝐷𝐴𝑣2 − 𝑣 Front of moving object
Rolling resistance 𝐹𝑅 𝐹𝑅 = 𝐶𝑅 ∙ 𝐹𝑁 − 𝑣 Contact surface
Buoyancy, Upthrust 𝐹𝑈 𝐹𝑈 = 𝜌 ∙ 𝑉 ∙ 𝑔 Along pressure gradient 𝛻𝑝 Lowest point of object
Tension 𝐹𝑇 Reaction force ∥ the cord End of the cord
Contact force 𝐹𝑃𝑢𝑠ℎ 𝐹𝑁 Reaction force ⊥ surface Contact surface
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Tension, Compression, Shear
Vectors in mechanics 25
Compression
Tension
Shear
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Resultant, Net, Overall force
Vectors in mechanics 26
𝐹𝑊
𝐹𝐴𝑖𝑟
Σ 𝐹
𝐹𝑁𝐸𝑇𝑇
𝐹𝑅𝐸𝑆
Σ 𝐹1st choice
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Linear momentum
Vectors in mechanics 27
𝑝 = 𝑚 𝑣 Momentum is conserved in an isolated system Σ 𝑝 = constant
𝑚𝐴 = 5.0 𝑔
𝑣𝐴 = 300 𝑚𝑠−1
𝑚𝐵 = 1.0 𝑘𝑔
𝑣𝐵 = 0
𝑣𝑒𝑛𝑑 ?
Σ𝑝𝑏𝑒𝑓𝑜𝑟𝑒 = Σ𝑝𝑎𝑓𝑡𝑒𝑟 ⇒ 𝑚𝐴 ∙ 𝑣𝐴 +𝑚𝐵 ∙ 𝑣𝐵 = 𝑚𝐴 +𝑚𝐵 ∙ 𝑣𝑒𝑛𝑑
5.0 ∙ 10−3 ∙ 300 + 1.0 ∙ 0 = 5.0 ∙ 10−3 + 1.0 ∙ 𝑣𝑒𝑛𝑑 ⇒ 𝑣𝑒𝑛𝑑 = 1.5 𝑚𝑠−1
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Newton’s laws
Vectors in mechanics 28
For each separate object:
Between objects:
Σ 𝐹 =Δ 𝑝
Δ𝑡Σ 𝐹 = 𝑚
Δ 𝑣
Δ𝑡⇒ Σ 𝐹 = 𝑚 𝑎
Constant mass2
If the resultant force on an object is zero, the object is in “translational equilibrium”1
3 ∆ 𝑝𝐴 = −∆ 𝑝𝐵 ⇒∆ 𝑝𝐴∆𝑡=∆ 𝑝𝐵∆𝑡
⇒ 𝐹𝐴𝐵 = − 𝐹𝐵𝐴
𝐹𝐵𝐴 𝐹𝐴𝐵
A B
“Force by A on B”
Wikimedia.org
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Newton’s laws – Example (1/5)
Vectors in mechanics 29
Neglect air resistance in this problem.
A caravan (820 kg) is pulled by a car. The combination accelerates with 𝑎 = 0.60 𝑚𝑠−2.The caravan is subject to a rolling resistance of 𝐹𝑟𝑜𝑙𝑙 = 1.2 ∙ 10
2𝑁
Question 1: Calculated the pulling force by the car on the caravan.
The car with passengers (total 1580 kg) is subject to a rolling resistance of 1.6 ∙ 102𝑁 during the acceleration.
Question 2: Calculate the forward force by the road on the car.
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Newton’s laws – Example (2/5)
Vectors in mechanics 30
A
B
𝑚𝐴 = 1580 𝑘𝑔
𝑚𝐵 = 820 𝑘𝑔
𝑎 = 0.60𝑚𝑠−2
+
Preparation: Make sketch of the connected car & caravan. Designate them ‘A’ & ‘B’.List the information given in the text: mass, acceleration and define a PLUS direction!
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Newton’s laws – Example (3/5)
Vectors in mechanics 31
A
B
𝑚𝐴 = 1580 𝑘𝑔
𝑚𝐵 = 820 𝑘𝑔
𝑎 = 0.60𝑚𝑠−2
+
𝐹𝐴𝐵
𝐹𝑟𝑜𝑙𝑙,𝐵 = −1.2 ∙ 102𝑁
1: Draw the relevant force vectors on the caravan
Question 1
2: Apply 2nd law on B
Σ 𝐹𝐵 = 𝑚𝐵 𝑎 ⇒ 𝐹𝐴𝐵 + 𝐹𝑟𝑜𝑙𝑙,𝐵 = 𝑚𝐵 𝑎 ⇒ 𝐹𝐴𝐵 + −1.2 ∙ 102 = 820 ∙ +0.60 ⇒ 𝐹𝐴𝐵 = 612𝑁
Round result to 6.1 ∙ 102𝑁
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Newton’s laws – Example (4/5)
Vectors in mechanics 32
A
B
𝑚𝐴 = 1580 𝑘𝑔
𝑚𝐵 = 820 𝑘𝑔
𝑎 = 0.60𝑚𝑠−2
+
𝐹𝐴𝐵 = 612𝑁
𝐹𝑟𝑜𝑙𝑙,𝐵 = −1.2 ∙ 102𝑁
1: Add the relevant force vectors on the car
Question 2
3: Apply 2nd law on A
Round result to 1.7 ∙ 103𝑁
𝐹𝐵𝐴
𝐹𝑟𝑜𝑙𝑙,𝐴 = −1.6 ∙ 102𝑁
𝐹𝑟𝑜𝑎𝑑,𝐴
2: Apply 3rd law on A: 𝐹𝐵𝐴 = − 𝐹𝐵𝐴 = −612𝑁
Σ 𝐹𝐴 = 𝑚𝐴 𝑎 ⇒ 𝐹𝑟𝑜𝑎𝑑,𝐴 + 𝐹𝐵𝐴 + 𝐹𝑟𝑜𝑙𝑙,𝐴 = 𝑚𝐵 𝑎 ⇒ 𝐹𝑟𝑜𝑎𝑑,𝐴 − 612 − 1.6 ∙ 102 = 1580 ∙ 0.60 ⇒ 𝐹𝑟𝑜𝑎𝑑,𝐴 = 1720𝑁
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Newton’s laws – Example (5/5)
Vectors in mechanics 33
A
B
𝑚𝐴 = 1580 𝑘𝑔
𝑚𝐵 = 820 𝑘𝑔
𝑎 = 0.60𝑚𝑠−2
+
𝐹𝐴𝐵
𝐹𝑟𝑜𝑙𝑙,𝐵 = −1.2 ∙ 102𝑁
1: Add the relevant force vectors on the car
Question 2: Alternative solution: treat A & B as combination AB
3: Apply 2nd law on A and B together
Round result to 1.7 ∙ 103𝑁
𝐹𝐵𝐴
𝐹𝑟𝑜𝑙𝑙,𝐴 = −1.6 ∙ 102𝑁
𝐹𝑟𝑜𝑎𝑑,𝐴
2: Regard the combination as a whole. Now 𝐹𝐴𝐵 and 𝐹𝐵𝐴 are internal forces and cancel out !
Σ 𝐹𝐴𝐵 = 𝑚𝐴 +𝑚𝐵 𝑎 ⇒ 𝐹𝑟𝑜𝑎𝑑,𝐴 + 𝐹𝑟𝑜𝑙𝑙,𝐴 + 𝐹𝑟𝑜𝑙𝑙,𝐵 = 𝑚𝐴 +𝑚𝐵 𝑎 ⇒
⇒ 𝐹𝑟𝑜𝑎𝑑,𝐴 − 1.2 ∙ 102 − 1.6 ∙ 102 = 1580 + 820 ∙ 0.60 ⇒ 𝐹𝑟𝑜𝑎𝑑,𝐴 = 1720𝑁
AB
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Impulse
Vectors in mechanics 34
𝐹 =∆ 𝑝
∆𝑡⇒ ∆ 𝑝 = 𝐹 ∙ ∆𝑡
𝑚 ∙ ∆ 𝑣 = 𝐹 ∙ ∆𝑡
𝑚 ∙ 𝑣2 − 𝑣1 = 𝑡0
𝑡1 𝐹 𝑡 ∙ 𝑑𝑡
Constant mass
𝐹
𝑡𝑡1 𝑡2
Constant force
Photo: Wikimedia.org
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Impulse – Example (1/2)
Vectors in mechanics 35
𝐹 𝑁
𝑡
1000
A baseball (145 g) is thrown and flies with 155 m/s. The hitter’s club hits the ball frontally. Assume that the ballFlies back in the opposite direction. The force by the club on the ball is drawn in the diagram below.
Question: Determine the speed after the hit.
105𝑚𝑠
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Impulse – Example (2/2)
Vectors in mechanics 36
𝐹 𝑁
𝑡
1000
95𝑚𝑠
3: ∆ 𝑝 = 𝐹 ∙ ∆𝑡 ⇒ 𝑚 𝑣2 − 𝑣1 = −42.3
2: The impulse 𝐹 ∙ ∆𝑡 equals the area under the graph. It can be approximated by the dashed triangle
𝐹 ∙ ∆𝑡 =1
2× 0.095 × −900 = −42.3𝑁𝑠
1: make sketch and choose PLUS direction 𝑣1 = +155𝑚𝑠
−1
𝑣2
∆ 𝑝 = 𝐹 ∙ ∆𝑡
Tock!
+ 𝐹
0.145 ∙ 𝑣2 − 155 = −42.3
𝑣2 = −140𝑚𝑠−1
4: Round to −1.4 ∙ 102𝑚𝑠−1
(graph area reading uncertainty)
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END
Vectors in mechanics 37
DisclaimerThis document is meant to be apprehended through professional teacher mediation (‘live in class’) together with a physics text book, preferably on IB level.