Determine the velocity of the car after it has been floored for 8 seconds

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Homework Help Overview

The discussion revolves around determining the velocity of a car after it has been accelerating for a specified duration. The problem involves concepts from dynamics, specifically the forces acting on a car, including thrust and drag, and requires an understanding of differential equations to solve for velocity over time.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the equation of motion related to the forces acting on the car and express varying levels of familiarity with the mathematical concepts involved. Some participants attempt to manipulate the equation to facilitate integration, while others seek clarification on the underlying calculus.

Discussion Status

The discussion is active, with participants exploring different mathematical approaches to the problem. Some guidance has been provided regarding the integration process, but there is no explicit consensus on the method or solution yet.

Contextual Notes

One participant notes a lack of familiarity with the required calculus, which may impact their ability to engage fully with the problem-solving process. There is also an emphasis on maintaining consistent units throughout the discussion.

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Hey if anyone could please tell me how to do this it would be a great help.

ok the question asks about an accelerating car. The mass of the car is 1200kg. When the engine is floored it creates a constant forward force of 9600N. The drag created by the car is equal to this equation: D=160v. Determine the velocity of the car after it has been floored for 8 seconds.
 
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You just need to solve the equation of motion:

[tex]m\frac {dv}{dt} = 9600 - 160 v[/tex]

Just remember to keep your units consistent!
 
actually i am not familiar with that equation. and my calculus is poor. So if anyone could explain this further, thanks!
 
[tex]1200\frac {dv}{dt} = 9600 - 160 v= 160(60- v)[/tex]
is the same as
[tex]\frac{dv}{60-v}= \frac{160}{1200}dt= \frac{1}{20}dt[/tex]

To integrate the left side, let u= 60- v so that du= -dv and the equation becomes
[tex]-\frac{du}{u}= \frac{1}{20}dt[/tex]

You ought to be able to integrate that.
 

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