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

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To calculate the constant acceleration needed for a car to accelerate from zero to 60 mph in 10 seconds, the acceleration 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 by the car during this acceleration can be calculated using the equation d = Vi(t) + 0.5(a)(t²), resulting in a distance of 134 meters or 440 feet. The discussion highlights the importance of using the correct equations for solving physics problems. Overall, the thread provides a step-by-step approach to understanding constant acceleration and distance calculations.
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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.
 
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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...
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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