Calculating acceleration when only have time and distance

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Homework Help Overview

The problem involves calculating the acceleration of a car that travels 400 meters over 60 seconds, with specific phases of acceleration, constant velocity, and deceleration. The context is kinematics, focusing on motion under varying conditions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate average velocity and seeks guidance on how to approach the problem. Some participants suggest starting with a graphical representation of velocity versus time, questioning the relationship between the area under the graph and displacement.

Discussion Status

Participants are actively engaging with the problem, exploring different methods of reasoning. Some guidance has been offered regarding the use of graphical analysis, and there is a recognition of the relationship between acceleration and distance in the context of the problem.

Contextual Notes

There are constraints related to the information provided, as the original poster does not have specific equations or values for cruising speed and is working with limited data. The discussion also reflects uncertainty about the calculations and assumptions made regarding the phases of motion.

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Homework Statement


A car travels 400m in 60s. The first 10s it accelerates from stationary. The last 10 seconds it decelerates back to stationary. For the middle 40s it has a constant velocity. What is the acceleration?

Homework Equations


None given

The Attempt at a Solution


Using only time and distance traveled I can only think how to calculate the average velocity: Vave = distance traveled / time = 6.67 m/s. I don't so much need an answer, more some guidance of where to begin. I'm pretty stuck. Thanks
 
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Maybe start graphically? You might find some inspiration. Make a sketch of velocity versus time. You don't know the cruising speed yet, so leave that as a variable. Now, what do you know about the area under a velocity vs time graph?
 
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The area under the graph should equal the displacement, so the total area should equal 400m. I have drawn a graph (attached). When a put the acceleration and deceleration together the area is exactly one 5th of the time spent cruising: therefore, the amount of time spent accelerating and decelerating is one 5th of 400m = 80 meters. Divide this by 2 and the car moved 40 meters in 10 seconds.
Acceleration = distance/time^2 = 40m/10^2 = 0.4m/s

Does this sound right? Making a sketch was good advise because it is starting to make sense to me. Just hope I have taken a wrong turn somewhere.
Thanks :)
 

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  • Time vs Velocity.png
    Time vs Velocity.png
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Very close indeed. However: ##distance = \frac{1}{2} Acceleration \cdot time^2##. So fix that up and you should be good.
 
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Yay! Thank you for your help. I've been staring at this question since yesterday, and actually drawing it was brilliant advice. Thank you!
 

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