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

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SUMMARY

The discussion focuses on calculating the constant acceleration required for a car to accelerate from 0 to 60 mph in 10 seconds, which is determined to be 2.68 m/s². This acceleration is approximately 27.3% of gravitational acceleration (g = 9.8 m/s²). The distance traveled during this acceleration is calculated using the equation d = Vi(t) + 0.5(a)(t²), resulting in a distance of 134 meters or 440 feet. The forum participant successfully applied the kinematic equation to derive these results, demonstrating a practical application of physics concepts.

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  • Understanding of kinematic equations in physics
  • Knowledge of unit conversions (mph to m/s and meters to feet)
  • Familiarity with the concept of constant acceleration
  • Basic grasp of gravitational acceleration (g = 9.8 m/s²)
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  • Learn more about kinematic equations and their applications in physics
  • Explore unit conversion techniques for speed and distance
  • Study the implications of constant acceleration in real-world scenarios
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Students studying physics, particularly those learning about motion and acceleration, as well as educators seeking practical examples for teaching kinematics.

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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:
<br /> x = \frac{1}{2}a t^2 + v_0 t + x_0<br />
 
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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