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The characteristics of the kinematics allow to find the results for the questions about the movement of the body are:
a) we have four sections;
- 0 to 8 s The body is accelerating.
- 8 to 10 s The body goes at a constant speed, the acceleration is zero.
- 10 to 14 Body accelerating.
- 14 to 18 Body slowing down.
b) The acceleration is the first 8 s is: a = 0.25 m / s²
c) The maximum acceleration is: a = 1 m / s²
d) The displacement is: i) d₁ = 8m, ii) = 16 m
e) maximum speed is: v = 6 m / s
Kinematics studies the movement of bodies by finding relationships between the position, speed and acceleration of bodies.
v = v₀ + a t
y = v₀ t + ½ a t²
where v and v₀ is the current and initial velocity, respectively, a is the acceleration and t is time.
In many circumstances graphs are made for their analysis, in a graph of speed versus time when we have a horizontal line the speed is constant, the acceleration is zero and in the case of a slope there is an acceleration, we have two cases:
- Positive slope the body is accelerating and the speed is increasing.
- Negative slope the body is stopping, the speed decreases.
Let's answer the different questions about the system.
a) in the attached we have a graph of the velocity versus time, each section corresponds to a change in the slope of the graph, we have four sections;
- 0 to 8 s The body is accelerating.
- 8 to 10 s The body goes at a constant speed, the acceleration is zero.
- 10 to 14 Body accelerating.
- 14 to 18 Body slowing down.
b) The acceleration is the first 8 s
v = v₀ + a t
a = 0.25 m / s²
c) The maximum acceleration is when the slope is maximum.
a = 1 m / s²
Therefore the acceleration is maximum in the section between 10 and 14 s
d) The total displacement is the sum of the displacements of each section.
We look for every displacement.
d₁ = v₀ + ½ a₁ Δt²
d₁ = 0 + ½ 0.25 8²
d₁ = 8 m
In the second section the velocity is constant
d₂ = v₂ Δt₂
d₂ = 2 (10-8)
d₂ = 4 m
The third section.
d₃ = v₀ + ½ a t²
d₃ = 2 + ½ 1 (14-10) ²
d₃ = 10 m
The distance of the fourth section.
we look for acceleration
a₄ =
a₄ =
a₄ = -1.5 m / s²
d₄ = 6 + ½ (-1.5) (1814) ²
d₄ = -6 m
The total displacement is;
= 8 + 4 + 10 -6
= 16 m
e) The maximum speed is the highest point in the graph of speed versus time that in the attachment we can see corresponds to
v = 6 m / s
In conclusion using the characteristics of kinematics we can find the results for the questions about the motion of bodies are:
a) we have four sections;
- 0 to 8 s The body is accelerating.
- 8 to 10 s The body goes at a constant speed, the acceleration is zero.
- 10 to 14 Body accelerating.
- 14 to 18 Body slowing down.
b) The acceleration is the first 8 s is: a = 0.25 m / s²
c) The maximum acceleration is: a = 1 m / s²
d) The displacement is: i) d₁ = 8m, ii) = 16 m
e) maximum speed is: v = 6 m / s
Learn more about kinematics here: brainly.com/question/24783036
Answer:
a) 138 units
b) 17 units
c) 17 units
d) Total Cost = $353.35
Explanation:
Given:
Average pizzas delivered = 200
Charge of inventory holding = 30% of cost
Lead time = 7 days
Now,
a) Economic Order Quantity =
also,
Annual Demand = 200 × 12 = 2400
Cost per Order = Cost of Box + Processing Costs
= 30 cents + $10
= $10.30
and, Carrying Cost =
=
=
= $2.575
Therefore,
Economic Order Quantity =
= 138.56 ≈ 138 units
b) Reorder Point
= (average daily unit sales × the lead time in days) + safety stock
= (
= 46.67 ≈ 47 units
c) Number of orders per year =
=
= 17.39 ≈ 17 units
d) Total Annual Cost (Total Inventory Cost)
= Ordering Cost + Holding Cost
Now,
The ordering Cost = Cost per Order × Total Number of orders per year
= $10.30 × 17
= $175.1
and,
Holding Cost = Average Inventory Held × Carrying Cost per unit
Average Inventory Held = = 69
Carrying Cost per unit = $2.575
Holding Cost = 69 × $2.575 =
$177.675
Therefore,
Total Cost = Ordering Cost + carrying cost
= $175.1 + $177.675 = $353.35