Model: Locutusque/Hyperion-3.0-Yi-34B
Author: Locutusque
Pipeline: text-generation
Library: transformers
Tags: transformers, safetensors, llama, text-generation, code, chemistry, medical, en, dataset:Locutusque/hyperion-v3.0, license:other, text-generation-inference, endpoints_compatible, region:us
Card metadata: language: en
license: other
library_name: transformers
tags: code, chemistry, medical
datasets: Locutusque/hyperion-v3.0
widget: example_title: Derivative
text: 
user

In this task, you are given a description of an object and its surroundings. Your
task is to describe the motion of the object mentioned in the description using
differential equations. The answer should be a mathematical expression in t. Description:
A ball is thrown up at 30 ft/sec from 3 feet off the ground. Solve this step-by-step,
explaining each step in detail.



assistant

output: text: Let's denote the height of the ball as h(t) where t represents time in seconds. Since the ball is thrown up with an initial velocity of 30 ft/sec, we can write the following equation for the vertical component of the ball's velocity:

v(t) = -32t + 30

Here, v(t) represents the velocity of the ball at time t (in feet per second), and the negative sign indicates that the velocity decreases due to gravity. Now, we can integrate the velocity function to find the position function:

h(t) = ∫v(t) dt = ∫(-32t + 30) dt

Integrating each term separately, we get:

h(t) = (-16t^2 + 30t + C)

To determine the constant C, we use the initial condition that the ball is thrown from 3 feet above the ground. At t=0, the height is 3 feet:

h(0) = 3

Substituting t=0 into the equation for h(t):

3 = (-16(0)^2 + 30(0) + C)

Solving for C, we get:

C = 3

So the final equation for the height of the ball as a function of time is:

h(t) = -16t^2 + 30t + 3

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Created: 2024-03-16 22:50:51