Please help, Resistive Forces, Initial Acceleration.

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Homework Help Overview

The problem involves calculating the initial acceleration of a sports car experiencing resistive forces due to aerodynamic drag as it coasts in neutral. The car's mass, drag coefficient, frontal area, and air density are provided, along with an initial speed converted from km/h to m/s.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the formula for resistive force and its implications for acceleration. There are attempts to equate resistive force with mass times acceleration, leading to questions about unit conversions and calculations. Some participants express uncertainty about the signs of their results and the implications of deceleration.

Discussion Status

The discussion is ongoing, with participants sharing their calculations and questioning the accuracy of their results. There is acknowledgment of potential errors in calculations, particularly regarding the squaring of velocity and unit conversions. Multiple interpretations of the resistive force and its impact on acceleration are being explored.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is a focus on ensuring correct application of the physics principles involved, particularly regarding resistive forces and their effects on motion.

yb1013
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Homework Statement



The mass of a sports car is 900 kg. The shape of the car is such that the aerodynamic drag coefficient is 0.250 and the frontal area is 2.00 m2. Neglecting all other sources of friction, calculate the initial acceleration of the car ,if it has been traveling at 100 km/h and is now shifted into neutral and is allowed to coast. (Take the density of air to be 1.295 kg/m2.)

Homework Equations



R=.5 *D*p*A*v^2

The Attempt at a Solution



I calculated it out and got that R is equal to 9, however I am not quite sure what to do with the Resistive Force.
 
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Doesn't F = m*a still?
 
I guess it would, so does that mean that i would just do 9=ma?
because i did that but then you get like .01 and that's not correct. Plus i know the answer has to be negative..
 
yb1013 said:
I guess it would, so does that mean that i would just do 9=ma?
because i did that but then you get like .01 and that's not correct. Plus i know the answer has to be negative..

Well of course it's decelerating.

Apply minus sign accordingly.
 
Btw, I didn't calculate your number. I suppose you converted your units correctly?
 
i believe i did,
the drag coeff. is still .25
the frontal area is 2 m^2
100 km/h converts to 27.8 m/s
and i think the density of the air is still 1.295 km/m^2

After i did that plugged them into the equation i got that the ressistance force is 17.99

so 17.99=(900) (a)?

a= .02

Where did i go wrong?
 
im sorry, density of air was kg/m^2
 
That would be different from your first answer of 9.

And it is (-) because the force is resistive to the direction of motion.
 
o I am sorry, i forgot to plug 1/2 in the beginning of the equation, so it really is 9, and then that divided by 900 equals .01.

but -.01 isn't the correct initial velocity..
 
  • #10
yb1013 said:
o I am sorry, i forgot to plug 1/2 in the beginning of the equation, so it really is 9, and then that divided by 900 equals .01.

but -.01 isn't the correct initial velocity..

You mean deceleration don't you?
 
  • #11
yes, that's what i meant, but do you have any idea what I am doing wrong?
 
  • #12
yb1013 said:
R=.5 *D*p*A*v^2

R = .5*(.25)(1.295)*(2)*(27.8)2

and that only comes out to 9?

I get a different answer.
 
  • #13
oo thank you, i seemed to forget to square the velocity for some reason?
but anyways, thank you very much!
 

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