Rearranging formulas. Accelerations

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To solve for average acceleration (Aav) in the equation Δd = viΔt + 1/2aavΔt^2, it's important to note that this formula applies only under constant acceleration. The average acceleration can be visualized as the slope on a velocity-time graph, while the distance traveled corresponds to the area under that curve. An example illustrates that maintaining a constant speed after initial acceleration can lead to a greater distance covered despite having the same average acceleration. Thus, understanding the relationship between acceleration and distance is crucial in physics. This highlights the complexity of motion beyond simple formulas.
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If I were to solve for Aav

Δd = viΔt+1/2aavΔt^2

Vi- initial velocity
Aav is average acceleration
 
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No, that only works for constant acceleration. Consider a velocity-time graph. The average acceleration is the slope from the start point to finish point, while the distance traveled is the area under the curve. E.g. if you stayed put for an hour then accelerated at 10kph/s for 6 seconds you'd average 1 kph/minute but cover relatively little ground. Conversely, if accelerate for the first 6 seconds then maintain a 60kph an hour you'll have the same average acceleration but go much further.
 
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