How to arrive at answer for integration question

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Homework Help Overview

The problem involves finding the antiderivative of acceleration for a stone thrown upwards, with a specified initial velocity and constant gravitational acceleration. The context is calculus, specifically integration and motion under gravity.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to find the antiderivative of the given acceleration but expresses uncertainty about their calculations. Some participants suggest focusing on the integral of the acceleration without units, while others clarify the correct value of acceleration to consider.

Discussion Status

Participants are engaging in clarifying the approach to the problem, with some guidance provided regarding the integration process. There is an acknowledgment of the need to correct the acceleration value, but no consensus on the final solution has been reached.

Contextual Notes

The original poster indicates they are a beginner and may be unsure about the appropriate section for their question. There is also a note that the constant of integration is not being solved for.

Alexenn
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Firstly, I am sorry if this is the wrong section (I really didn't know what each section meant, but I posted here since I am beginner)

Homework Statement



A stone is thrown up into the air at 20m/s is accelerated constantly downwards by gravity at about 10m/s2

Find the anti derivative of the acceleration

Note: I am not solving for C

Homework Equations


The Attempt at a Solution



The antiderivative of 20m/s is 10m/s2
and the antiderivate of -10m/s is -5m/s2

Neither of those match up to the answer, so I'm not sure how to arrive at the answer.

The answer in the book is v= -10t + c

Thankyou
 
Last edited:
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If you have [itex]a = 10[/itex] then you need to find [itex]\int 10\ dt[/itex]. There's no need to include units in your integral.
 
Ah, thank you very much. Can finally get on with work! =)
 
JHamm said:
If you have [itex]a = 10[/itex] then you need to find [itex]\int 10\ dt[/itex]. There's no need to include units in your integral.
That should be [itex]a = -10[/itex]. Of course, the units are m/s2 .
 

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