UBER EASY QUESITON: just need to make sure

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SUMMARY

The discussion centers on calculating average velocity and acceleration for an object in free fall, assuming negligible air resistance. The user correctly identifies that average velocity can be calculated using the formula d/t, where d is distance and t is time. It is established that while the average velocity is reached at half the total time, the object's speed continues to increase due to constant acceleration. The relationship between distance, acceleration, and time is clarified with the formula d = 1/2 a t², confirming that acceleration can be derived from the known variables.

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pfssassin
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Just a general question. If you drop a object (air resistance negligible) and you know the t initial is 0 and you know the t final is w/e you get on your stop watch. You do d/t to get average velocity of the fallen object.


Now that you have the avrg velocity you know the V initial is zero and the speed rises to the avrg velocity, can you assume that a = Vavrg/t?


Im in IB physics HL but just can't get my head around this one for some reason...
 
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Why do you think the speed only rises to the average velocity? If you assume constant acceleration and you had a way to measure the final velocity, then you would be right. But the average velocity would be reached at 1/2 t_final and keep increasing after that...

Out of the data you have (t and d) you can find out a by using [tex]d = \frac{1}{2} a t^2[/tex]
 
haha wow I feel like an idiot. Thanks.
 

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