Calculating acceleration when only have time and distance

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A car travels 400 meters in 60 seconds, accelerating for the first 10 seconds and decelerating for the last 10 seconds, with a constant velocity in between. The average velocity is calculated at 6.67 m/s, but the focus is on determining the acceleration. By sketching a velocity versus time graph, it is understood that the area under the graph represents displacement, leading to the conclusion that the car covers 80 meters during acceleration and deceleration. This results in an acceleration of 0.4 m/s², although a correction is suggested to use the formula for distance during acceleration. The discussion emphasizes the importance of visual aids in solving physics problems.
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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


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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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