How Do You Calculate Distance Traveled by a Car Under Constant Acceleration?

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Homework Help Overview

The discussion revolves around calculating the distance traveled by a car under constant acceleration, specifically transitioning from zero to 60 mph in a set time frame. The problem involves concepts from kinematics and acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to solve for acceleration and distance using kinematic equations, expressing confusion about the distance calculation. Other participants suggest using standard equations from textbooks and confirm the equation used by the original poster.

Discussion Status

The discussion has progressed with the original poster reporting a successful resolution to part c of the problem. Participants have engaged in confirming the approach and sharing encouragement, although no explicit consensus on the method has been established.

Contextual Notes

The original poster mentions working under a time constraint and expresses a desire for clarity, indicating a learning environment where assumptions and understanding are being actively questioned.

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[SOLVED] Constant Acceleration

Homework Statement


a. What constant acceleration, in SI units, must a car have to go from zero to 60 mph in 10s?

b. What fraction of g is this?

c. How far has the car traveled when it reaches 60 mph? Give your answer both in SI units and in feet.


Homework Equations





The Attempt at a Solution


I have gotten answers for parts a and b, but I get really confused on part c.

For a: 1 mph = .447 m/s, so 60(.447)=26.82 m/s. (26.82 m/s)/(10s)=2.68 m/s^2

For b: g=gravity=9.8 m/s^2. (2.68 m/s^2)/(9.8 m/s^2)=.273(100)=27.3%

For c: I'm not sure where to start to find my answer.

I appreciate any help!

I just solved it! I've worked on part C for about 45 minutes to an hour. I just found an equation that worked, imagine that.

For c: I used d=Vi(t)+0.5(a)(t^2) So, d=(0 m/s)(10 s)+0.5(2.68 m/s^2)(10^2)=134 m. Then to convert 134 m to feet = 134(39.37)=5,275.58 in/12=440 ft

Hopefully this will help someone else out that is new to physics, like myself.
 
Last edited:
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use one of them there equations from your textbook:
[tex] x = \frac{1}{2}a t^2 + v_0 t + x_0[/tex]
 
This is the exact equation I used! Please see the bottom of my previous post, I just finished editing my initial post. I appreciate your help, all of the great help I've received from this forum!
 
ah. well, that's good.
 
Yeah and hopefully your teacher can help... oh wait...
 

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