How Do You Calculate Average Acceleration in a Piston Cycle?

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To calculate average acceleration in a piston cycle with a displacement of 10 cm and a frequency of 50 cycles per second, the average velocity is determined using the formula v = d / t. The time for the piston to travel 5 cm is calculated as 5 ms, leading to the acceleration formula a = 2 * d / t². The correct calculation requires squaring the time value, resulting in a = 2 * 0.05 m / (0.005 s)², which equals 20 m/s². Clarification is provided that the time variable must be squared in the formula for accurate results.
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1. Homework Statement
The piston has a displacement of 10 cm and there are 50 cycles per second. Calculate the average acceleration that occurs between the start/stop point and the middle of the cycle. This value has to do with evenly accelerated movement. This can be described according to the formula d = .5a * t2.
To solve:
-Compute the average velocity using a full cycle.
-Calculate how long the piston needs to travel 5 cm from the top of the cycle to the middle, which will give us the value for t. Then use a = 2 * d / t2 to calculate the acceleration.

2. Homework Equations
d = .5a * t2
a = 2 * d / t2
v = d / t

3. The Attempt at a Solution
50 cycles per second = 1 cycle takes 20 milliseconds.
velocity = d / t
10 m/s = 20 cm / 50 ms (total cycle distance is 20cm because the displacement is 10cm)
How long the piston takes to travel 5 cm: t = 5 ms
//But why figure out how long 5 cm takes, when I'm supposed to be solving for the middle of the cycle, which is 10 cm?
//Solving for 5 cm:
a = 2 * d / t2
d = 5 cm = .05 m
t = 5 ms = .005 s
a = 2 * .05 m / .005s2 = 20 m/s2
//But this isn't the right answer... How does the value for t work if it's supposed to be in seconds squared instead of just seconds? Should I plug in .0052 instead of just .005? Help!
 
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After a full cycle the piston is back where it started from. Displacement from its point of origin is zero. So what's its average velocity?

If a formula says the value is to be squared, then yes, when you plug in the value for the variable it is to be squared. The "value" of the variable includes its units, so if the formula calls for t2 and t is 5s, then t2 is going to be 25s^2.
 
Palladian1881 said:
2. Homework Equations
d = .5a * t2
a = 2 * d / t2
Note that these equations should be:
d = (.5)a * t2
a = 2d/(t2)


Palladian1881 said:
//Solving for 5 cm:
a = 2 * d / t2
d = 5 cm = .05 m
t = 5 ms = .005 s
a = 2 * .05 m / .005s2 = 20 m/s2
//But this isn't the right answer... How does the value for t work if it's supposed to be in seconds squared instead of just seconds? Should I plug in .0052 instead of just .005? Help!


You forgot to square the value for t.
 
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