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@@ -5,60 +5,88 @@ language:
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  - en
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  widget:
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  - example_title: Reasoning
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- text: |
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  <|im_start|>system
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- You are Hyperion, an AI assistant developed by Sebastian Gabarain with a focus in reasoning, coding, science, and other general-purpose subjects. You are an AI assistant that follows instruction extremely well. Help as much as you can.
 
 
 
 
 
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  <|im_end|>
 
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  <|im_start|>user
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- In this task, you are given a description of an object and its surroundings. Your
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- task is to describe the motion of the object mentioned in the description using
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- differential equations. The answer should be a mathematical expression in t. Description:
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- A ball is thrown up at 30 ft/sec from 3 feet off the ground. Solve this step-by-step,
 
 
 
 
 
 
 
 
 
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  explaining each step in detail.
 
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  <|im_end|>
 
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  <|im_start|>assistant
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-
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  output:
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- text: |
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- Hyperion here. The task requires you to describe the motion of a ball thrown up using differential equations. Here's the detailed explanation of how to solve it:
 
 
 
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  1. Identify the given information:
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  - Initial velocity (v0): 30 ft/sec
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  - Initial height (y0): 3 feet off the ground
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-
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- 2. Determine the acceleration due to gravity (g) at sea level, which is approximately -32.2 ft/s^2. Since we're dealing with a short distance and not very high altitudes, we can consider this value of g as constant.
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-
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- 3. Use Newton's second law of motion to find the differential equation that describes the motion:
 
 
 
 
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  - The net force acting on the ball is the sum of gravity (mg) and air resistance (which we will neglect for simplicity). So, Fnet = mg.
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  - Since acceleration equals force divided by mass, a = Fnet/m. In this case, m cancels out, so a = g.
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  - The motion equation becomes: dv/dt = g = -32.2 ft/s^2 (since we're moving upwards). This is the first differential equation.
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-
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- 4. Since we want to describe the ball's position (y) as well, we need another differential equation. We can use kinematics equations:
 
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  - The velocity equation is v = v0 + at. Substitute the value of a from step 3:
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  dv/dt = -32.2 ft/s^2. Rewriting it as an integral:
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  ∫dv = -32.2 ft/s^2 * dt
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  - Integrating with respect to time, we get v(t) = -16.1 t + C (constant of integration).
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-
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  5. Now we have two equations, one for velocity and another for position:
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  - dv/dt = -32.2 ft/s^2 (equation 1)
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  - v(t) = -16.1 t + C (equation 2)
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-
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- 6. We need to find the constant C using the initial conditions given in the problem. We know that initially, at time t=0, the ball's velocity is v0:
 
 
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  - Equation 2 becomes v0 = -16.1 * 0 + C, which simplifies to C = v0 = 30 ft/sec (initial velocity).
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-
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  7. Now we have both equations with the constants filled in:
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  - dv/dt = -32.2 ft/s^2 (equation 1)
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  - v(t) = -16.1 t + 30 ft/sec (equation 2)
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-
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  8. Finally, we need to find the ball's position (y) as a function of time:
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  - Use the second kinematics equation y = y0 + v0*t + ½*a*t^2 (since acceleration is constant, it can be moved outside the integral).
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  - Substitute the values for v0 and a:
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  y(t) = 3 ft + (30 ft/sec) * t + (-½ * 32.2 ft/s^2) * t^2
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-
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  That's it! The differential equation describing the motion of the ball is:
59
  - dv/dt = -32.2 ft/s^2 (equation 1)
60
  - v(t) = -16.1 t + 30 ft/sec (equation 2)
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  - y(t) = 3 ft + (30 ft/sec) * t - (16.1 t^2) ft (equation 3)
 
 
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  ---
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  # Hyperion-3.0-Mistral-7B-DPO
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5
  - en
6
  widget:
7
  - example_title: Reasoning
8
+ text: >
9
  <|im_start|>system
10
+
11
+ You are Hyperion, an AI assistant developed by Sebastian Gabarain with a
12
+ focus in reasoning, coding, science, and other general-purpose subjects. You
13
+ are an AI assistant that follows instruction extremely well. Help as much as
14
+ you can.
15
+
16
  <|im_end|>
17
+
18
  <|im_start|>user
19
+
20
+ In this task, you are given a description of an object and its surroundings.
21
+ Your
22
+
23
+ task is to describe the motion of the object mentioned in the description
24
+ using
25
+
26
+ differential equations. The answer should be a mathematical expression in t.
27
+ Description:
28
+
29
+ A ball is thrown up at 30 ft/sec from 3 feet off the ground. Solve this
30
+ step-by-step,
31
+
32
  explaining each step in detail.
33
+
34
  <|im_end|>
35
+
36
  <|im_start|>assistant
 
37
  output:
38
+ text: >
39
+ Hyperion here. The task requires you to describe the motion of a ball
40
+ thrown up using differential equations. Here's the detailed explanation of
41
+ how to solve it:
42
+
43
 
44
  1. Identify the given information:
45
  - Initial velocity (v0): 30 ft/sec
46
  - Initial height (y0): 3 feet off the ground
47
+
48
+ 2. Determine the acceleration due to gravity (g) at sea level, which is
49
+ approximately -32.2 ft/s^2. Since we're dealing with a short distance and
50
+ not very high altitudes, we can consider this value of g as constant.
51
+
52
+
53
+ 3. Use Newton's second law of motion to find the differential equation
54
+ that describes the motion:
55
  - The net force acting on the ball is the sum of gravity (mg) and air resistance (which we will neglect for simplicity). So, Fnet = mg.
56
  - Since acceleration equals force divided by mass, a = Fnet/m. In this case, m cancels out, so a = g.
57
  - The motion equation becomes: dv/dt = g = -32.2 ft/s^2 (since we're moving upwards). This is the first differential equation.
58
+
59
+ 4. Since we want to describe the ball's position (y) as well, we need
60
+ another differential equation. We can use kinematics equations:
61
  - The velocity equation is v = v0 + at. Substitute the value of a from step 3:
62
  dv/dt = -32.2 ft/s^2. Rewriting it as an integral:
63
  ∫dv = -32.2 ft/s^2 * dt
64
  - Integrating with respect to time, we get v(t) = -16.1 t + C (constant of integration).
65
+
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  5. Now we have two equations, one for velocity and another for position:
67
  - dv/dt = -32.2 ft/s^2 (equation 1)
68
  - v(t) = -16.1 t + C (equation 2)
69
+
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+ 6. We need to find the constant C using the initial conditions given in
71
+ the problem. We know that initially, at time t=0, the ball's velocity is
72
+ v0:
73
  - Equation 2 becomes v0 = -16.1 * 0 + C, which simplifies to C = v0 = 30 ft/sec (initial velocity).
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+
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  7. Now we have both equations with the constants filled in:
76
  - dv/dt = -32.2 ft/s^2 (equation 1)
77
  - v(t) = -16.1 t + 30 ft/sec (equation 2)
78
+
79
  8. Finally, we need to find the ball's position (y) as a function of time:
80
  - Use the second kinematics equation y = y0 + v0*t + ½*a*t^2 (since acceleration is constant, it can be moved outside the integral).
81
  - Substitute the values for v0 and a:
82
  y(t) = 3 ft + (30 ft/sec) * t + (-½ * 32.2 ft/s^2) * t^2
83
+
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  That's it! The differential equation describing the motion of the ball is:
85
  - dv/dt = -32.2 ft/s^2 (equation 1)
86
  - v(t) = -16.1 t + 30 ft/sec (equation 2)
87
  - y(t) = 3 ft + (30 ft/sec) * t - (16.1 t^2) ft (equation 3)
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+ datasets:
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+ - Locutusque/hyperion-dpo-v1.0
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  ---
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  # Hyperion-3.0-Mistral-7B-DPO
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