Sports Car Acceleration Question

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Homework Help Overview

The discussion revolves around a problem involving the acceleration of a sports car that initially travels at a constant speed and then brakes to a stop. The problem requires participants to determine the car's acceleration in terms of "g's."

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to find acceleration, with some questioning the meaning of variables such as 'u' and others clarifying the initial and final velocities involved in the calculations.

Discussion Status

The discussion includes attempts to identify the appropriate kinematic equations and clarify variable definitions. Some participants have provided guidance on which equations to consider, while others have worked through the calculations, leading to a numerical result.

Contextual Notes

There is mention of the need to express the acceleration in terms of "g's," which introduces a conversion factor into the discussion. The original poster expresses uncertainty about how to start the problem, indicating a potential gap in understanding the initial setup.

Adrianna
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Homework Statement


A sports car moving at constant speed travels 110m in 5.0s. If it then brakes and comes to a stop in 4.0s. what is its acceleration in m/s^2? Express the answer in terms of "g's." where 1.00 g = 9.80 m/s^2.


Homework Equations


I thought about using the equation a=v/t

The Attempt at a Solution


So I tried it sort of I am just sort of lost I am not really sure how to start...
 
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Adrianna said:

Homework Equations


I thought about using the equation a=v/t

You need to use the kinematic equations

v=u+at

v2=u2+2as

s=ut+1/2at2


Which one do you think you will have to use?
 
What does the u stand for, that's not in my book?
 
The car initially had a velocity of 110m/5s (22m/s) and came to a stop (0m/s) in 4.0s. The acceleration is the change in velocity over a time interval.
 
So would we use the v=u+at since I have the 4.0s and the 22m/s? but what would go in for the u?
 
u is initial velocity and v is final velocity.
 
Okay I figured it out thanks so much the answer was -5.5 m/s^2 and then in g's it was -0.56 g's
 
take u (initial speed) = 110 / 5 = 22m/s
v = 0
t = 4

v = u + at
0 = 22 + 4a
4a = -22
a = -5.5m/s^2
a = (-5.5/9.8)g = -0.56g
 

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