UBER EASY QUESITON: just need to make sure

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In the discussion, a user inquires about calculating average velocity and acceleration for a dropped object, assuming negligible air resistance. They initially suggest using average velocity to find acceleration, but others clarify that average velocity is reached at half the total time, indicating that speed continues to increase. It is emphasized that for constant acceleration, the relationship d = 1/2 a t^2 can be used to determine acceleration from distance and time. The user expresses realization about their misunderstanding and thanks the contributors for their help. This highlights the importance of understanding the relationship between average velocity and acceleration in physics.
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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 d = \frac{1}{2} a t^2
 
haha wow I feel like an idiot. Thanks.
 

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