Calculating Acceleration and Force in a Racecar: A Newtonian Approach

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Homework Help Overview

The problem involves calculating the acceleration and force experienced by a racecar covering a quarter-mile track in a specified time, assuming constant acceleration. The context is rooted in Newtonian mechanics, particularly focusing on concepts of force and acceleration.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to use kinematic equations to find acceleration and horizontal force but expresses uncertainty about the equations and their application. Some participants question the directions of forces involved and the assumptions made in the calculations.

Discussion Status

Participants are actively engaging with the problem, providing feedback on the original poster's attempts and clarifying the importance of force directions. There is a recognition of differing interpretations regarding the acceleration value, with some suggesting it may be expressed in terms of "g's." Guidance has been offered regarding the setup of equations.

Contextual Notes

There is mention of missing information regarding force directions, which has led to confusion in the calculations. The discussion reflects varying understandings of the problem setup and the implications of the given data.

yandereni
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Homework Statement


Here's the problem:
A particular racecarcan cover a quartermile track (402m) in 6.40s starting from standstill. Assuming that the acceleration is constant, how many "g's"(all i know is that it means acceleration due to gravity) does the driver experience? If the combined mass of the driver and the racecar is 485kg, what horizontal force must the road exert on the tires?

Homework Equations


1. Δy(displacement) = [(Velocityinitial)(Δt)] + ((1/2)acceleration)(Δt)2
2. Forcegrav + (acceleration)(mass)= Forcetotal
I'm not even sure if the second equation even exists but that's what i made out from the problem.

The Attempt at a Solution


I tried solving for the "g's" using the first equation and i substituted the possible things:

402m = [(0m/s)(6.4s)] + [((1/2)a)(6.4s)]
i don't know if this is the right equation but its the closest that i can get

then i tried solving for the horizontal force with the second equation and substituted the given:
[(-9.8m/s2)(485kg)] + [(19.6m/s2)(485kg)] = Forcetotal

and in case oyu are wondering where i got the acceleration(19.6m/s2) its from the first equation i solved (which i think is wrong)

Thank you in Advance!

-yandereni
 
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Hi, yandereni.

You did well in the first part. It looks correct.

yandereni said:
then i tried solving for the horizontal force with the second equation and substituted the given:
[(-9.8m/s2)(485kg)] + [(19.6m/s2)(485kg)] = Forcetotal
What are the directions of the two forces you've included here? What is the direction the question is concerned with?
 
I
Bandersnatch said:
Hi, yandereni.

You did well in the first part. It looks correct.What are the directions of the two forces you've included here? What is the direction the question is concerned with?
Bandersnatch said:
Hi, yandereni.

You did well in the first part. It looks correct.What are the directions of the two forces you've included here? What is the direction the question is concerned with?
I looked at the problem again and there were no directions stated. And when i told my other classmates about the first part if i did it right, they said no and the answer should be 2.00m/s2.
 
yandereni said:
I looked at the problem again and there were no directions stated.
what's this then, eh?
yandereni said:
what horizontal force must the road exert on the tires?

yandereni said:
when i told my other classmates about the first part if i did it right, they said no and the answer should be 2.00m/s2.
That's wrong. The car would cover some 41m at that acceleration.
 
Maybe they meant 2g, not 2m/s^2?
2g is about 19.62m/s^2.
 
oh, now i understand. but can i solve the horizontal force without any directions given?
 
yandereni said:
without any directions given?
But you do know which direction the car is moving, don't you? All you need to make sure when setting up your equation is not to include forces that are not acting (i.e., have no component) in the direction you're interested in.
 
Thank you very much! I got the right answer and i get the concept now. Thanks a lot! :)
 

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