# Free fall Experiment Help!!!!!

by bilalbajwa
Tags: experiment, fall, free
 P: 557 For #1 use the fact that acceleration is the rate at which velocity is changing. Since it is a constant acceleration, can you use a bit of math to find out how much the velocity increases in m/s in any given second? edit: woops sorry bilalbajwa I assumed that the 'objectives' were the questions, my bad. Let me look at the tables...
 P: 28 Q1. Average Acceleration = delta x/ delta ok i understand but what is yi/ti^2??? WHY yi/ti^2 is half of avg acceleration??? WHats the logic comming out of it???
P: 557

## Free fall Experiment Help!!!!!

The yi/ti^2 is the vertical displacement divided by the time interval squared.

Basically what is happening is when they are spitting out the 'velocity' they are measuring the instantaneous velocity! When you are looking at yi/ti you are looking at the AVERAGE velocity over that specific time interval. I will suppy some pictures in a second to help.
 P: 557 Actually I don't think you need pictures to understand this, I have to go shopping as well(will be back soon). But plot the average velocities yi/ti for each interval on the same graph as the recorded instantaneous velocities, what do you notice about the slope of each?
 P: 557 I am trying to write quick while making this as clear as possible. What did I show you in the other thread? You can use the midpoint rule for a line to derive an equation for the average velocity when you are dealing with constant acceleration. It is 1/2(vi+vf) right? So basically at the end of the experiment, if you look at the average velocity, it is only half of the actual final velocity and if you use that to compute the acceleration, you end up with 1/2 the actual acceleration, get it?
 P: 28 OK one thing i got here: SECTION I Coloumn comparison "Velocity" vs yi/ti "Velocity" = Instantaneous Velocity yi/ti = Average Velocity Conclusion: In a free falling body, when we take equal time intervals and unequal displacements. Avg Velocity is half of Instantaneous velocity??? AM I RIGHT IF YESY WHY IS IT SO???
 P: 557 It is due to the fact that there is a constant acceleration, not specifically because it is a 'free falling body'. Can you tell me why avg velocity is 1/2 of the 'instantaneous velocity at the END OF THE INTERVAL.'?
 P: 28 Because at the end of the interval Instantaneous velocity is on the peak and AVG velocity is the averages of the inital + final. (i am not sure)
 P: 557 I think you are getting there and probably understand. But anyways, the AVERAGE velocity of an object over a specific time interval (under constant acceleration) is $$v_{ave}=\frac{v_{i}+v_{f}}{2}$$ Since Vi is 0 in your the case of your object, you could express $$v_{ave}=\frac{v_{f}}{2}$$ Another way of expressing $$v_{ave}$$ is $$v_{ave}=\frac{\Delta d}{\Delta t}$$ Where delta d is the change in displacement and delta t is the change in time So this is where the yi/ti comes from, its simply the average velocity. I feel that I am not doing a very good job of explaining this.. :/

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