Accelerating a Car: Min Time to Reach Speed v

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Homework Help Overview

The problem involves a car's acceleration from rest to a specified speed, focusing on the relationship between power, work, and time. The subject area includes concepts from mechanics and energy considerations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between power, work, and time, with attempts to derive an equation for minimum time. Questions arise regarding the correctness of the final equation and the role of kinetic energy in the context of work done.

Discussion Status

Participants are actively exploring different approaches to the problem, including the use of kinetic energy versus work done. There is recognition of a potential error in the original equation, with suggestions for clarification regarding the factor of 1/2 in the context of energy considerations.

Contextual Notes

There is a mention of missing factors in the calculations and the need for average values in the context of work and energy. The discussion reflects on the definitions and relationships between power, force, and energy in the problem setup.

Kyoma
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Homework Statement



A car of mass m has an engine which can deliver power P. What is the minimum time in which the car can be accelerated from rest to a speed v?


2. The attempt at a solution

P = W/t
(W is work done, t is time)

F = ma

So,
P = Fs/t
(F is force, s is distance)
P = mas/t
Making t as the subject,
t = mas/P
t = mv2/P

Is there any thing wrong with my final equation?
 
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I am missing a factor of 1/2. Probably you forgot to take an average somewhere...
An easier way might be from an energy consideration: the amount of work that the engine needs to do, is at least the added kinetic energy, right?
 
If I use kinetic energy, instead of work done,

then the equation becomes:

t = mv2/2P

Which is the correct answer then? And what factor 1/2?
 
Exactly. Note that there is an extra 2 in the denominator now. That's the factor of 1/2 I am talking about.

(By the way, it is not "instead of work done"... the kinetic energy is precisely the work done. You meant, "instead of forces" :))
 

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