Calculating Particle Deceleration and Force | MCAT Physics Question 17

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To calculate the deceleration of a particle, the equation v(final)^2 = v(initial)^2 + 2ax was used, resulting in an initial value of -240000 for acceleration. However, a correction was provided, indicating that the correct deceleration should be -5.76 x 10^10 m/s^2. The force was then calculated using F = ma, yielding a force of 1.608 x 10^-21 N based on the initial acceleration. Participants in the discussion emphasized the importance of careful calculation and double-checking inputs. Accurate calculations are crucial for solving physics problems effectively.
accol
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1. http://www.prep101.com/MCAT/102MCATPhysicsAnswers.pdf number 17


3. Here is my attempt:

First I figure I have to find the deceleration of the particle. So I plugged it into the v(final)^2= v(initial)^2 + 2ax. I solved for a and got a and got -240000.

I don't know if I am correct here but I plugged the accel and mass into F = ma and got a force of 1.608*10^-21N

I also tried substuting equations such as =

F = Eq
to ma = Eq but I still haven't gotten an answer along the lines of the choices. Please help me.

Thanks in advance.
 
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accol said:
1. http://www.prep101.com/MCAT/102MCATPhysicsAnswers.pdf number 17


3. Here is my attempt:

First I figure I have to find the deceleration of the particle. So I plugged it into the v(final)^2= v(initial)^2 + 2ax. I solved for a and got a and got -240000.


You have the right idea, but I get a=-5.76 \times 10^{10} \text{m/s}^2. Double check your calculation :smile:
 
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Lol, wow...and I thought I was doing it all wrong, I just put the wrong numbers into the calculator. Thanks
 
Welcome :smile:
 
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