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The velocity is constantly increasing due to a constant acceleration. Exactly at 1 s the velocity is 10 m/s, but this does not mean that velocity was at 10 m/s in preceding second. In fact, given the distance 5 m moved in this second, the average velocity in this second was 5 m/s. And this should make sense to you, because in this first second the velocity ...

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You are right in that gravity did not change during data collection. You are a victim of uncertainty, which is a very important part of experimental physics. I'm sorry in advance for the "wall of text", and I hope that this clears up some confusion. The problem is that $1.50$ may not be exactly $1.500000000...$. Because the numbers are provided rounded, ...

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I agree with kevinsa5 that the variation in $a$ is due to rounding errors, but I'd like to suggest a better way to analyse the data. Generally speaking, the best way to analyse data is to find a way to convert it to a straight line, then you can graph it and do a linear regression. In this case the way to procede is to note that if the acceleration is ...

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Take the velocity vector a two points separated by an infinitesimally small time interval. Subtract the later vector from the earlier, and divide by the infinitesimally small time interval. The resulting vector is the instantaneous acceleration (in the limit that the time interval goes to zero). The direction of that vector is the direction of the ...

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This is correct, assuming constant acceleration, we have for this problem $$a = -\frac{v^2_i}{2d} = -\frac{(6.26)^2}{0.05} = -784\frac{m}{s^2}$$ First, I applaud you for asking the question. Too often, I have graded homework and tests where numbers were submitted for answers without any thought as to whether they were reasonable. However, this is a ...

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Well, let's think about this: An object is traveling at 6.26 m/s during impact and travels only 0.025 meters before stopping. The force causing the object to decelerate needs to be extremely high. It's just like having a force being applied for a very short period of time, such as a bat hitting a baseball and the time of contact is extremely small, you ...

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What is angular speed? Clearly it is $\frac {v_\perp} {r}$ where symbols have their usual meanings. Rod rotates about its, say, rightmost point, say $O$. We will consider left side as positive $x$-axis. Now consider a point $A$ at distance $r_1$ from it. Let the rod have instantaneous angular speed $\omega$. All points on the rod will have this $\omega$ ...

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Let's take a look just from the point of view of someone reading the problem. First of all, we can say "our car has motion", because it's changing its position each second. Ok. So, how is its motion? Well, it is moving in 1 dimention, it is a linear movement. Then, we can say "our car has linear motion". Also, we can see our car's velocity is changing ...

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The car probably experiences a constant acceleration of $10{m/s^2}$. You can see from the chart that the velocity follows this as after every second the car is going $10{m/s^2}$ faster. However, this is clearly not the whole picture. We do not know the acceleration at 1.5 seconds, or 1.55 or 3.14. We can get confirmation that our acceleration model works ...

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