Did I do this right an F=ma need to find Initial Speed?

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Homework Help Overview

The discussion revolves around a physics problem involving the application of Newton's second law (F=ma) to determine the initial speed of a car given its mass, braking force, and stopping distance. The original poster presents their calculations and seeks validation of their approach and results.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the initial speed using the given mass, force, and stopping distance, but questions arise regarding the method used to find time. Some participants point out that the formula applied for time requires the initial velocity, leading to confusion about the correctness of the calculations.

Discussion Status

Participants are actively engaging in clarifying the approach taken by the original poster. There is a recognition of the potential mistake in the time calculation, and some participants suggest that the negative value for initial velocity may be acceptable depending on the defined direction of motion. The discussion is ongoing, with multiple interpretations being explored.

Contextual Notes

There is a mention of the vector nature of velocity and acceleration, which may influence how initial velocity is interpreted in terms of sign. The original poster's calculations are based on specific assumptions about directionality, which are being questioned by others in the thread.

bradycat
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Did I do this right an F=ma need to find Initial Speed?

Need help on this.
A car whose mass is 2600 kg can produce an unbalanced braking force of 2765N.
Calculate the car's initial speed in order to have a stopping distance of 3036.3m.
Vi=?
So...
m=2600kg
F=2765N
s=3036.3m
Vf=0

So I found acceleration of a=F/m = 2765/2600 = 1.06346m/s^2
So I have to find time=Square root of 2(3036.33)/1.06346 = 75.56605s
So I would use this formula to solve for Vi? right
Vi=Vf -at
Vi = 0 - (1.06346)(75.56605) = -80.3614 m/s
Is the negative right?
IS THIS CORRECT if not please show me where I went wrong
Thanks
Jo
 
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Where you have found the time you have made a mistake since that formula requires the initial velocity too.
 


Kurdt said:
Where you have found the time you have made a mistake since that formula requires the initial velocity too.

Hi Kurdt,
I used the time formula where you use distancex2 over acceleration. THAT is wrong your saying?
What formula should I use?
 


Yes, that's wrong. It came from d=v0t + (1/2)at^2; you only get t=sqrt(2d/a) if v0=0, which, in this case, it's not. However, you got the right answer at the end because acceleration is symmetrical with respect to time. That is, if you record a car decelerating to 0 at -1.06 m/s^2 and play the video backwards at 1 second per second, the car would seem to be accelerating at 1.06 m/s^2 to its initial speed. That's why using t=sqrt(2d/a) worked.
 


Hi Ideasule.
So the -80.3614 m/s is correct?
So the Vi can have a negative number then?
I thought it could not have a negative on a Vi?
 


Velocities and accelerations are vector quantities, so they can certainly be negative. For Vi, yes, -80.36 m/s. If you don't explain the fact about time symmetry, however, you wouldn't get full credit on a test.

Whether Vi is positive or negative depends on which way you define as positive. Here, you used a=1.06 m/s^2, which means you've defined "backwards" as positive. It's not surprising, then, that initial velocity would be negative. (Initial SPEED, however, is not; speed is a scalar.)
 

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