What is the acceleration of the spore during the launch and speed reduction?

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SUMMARY

The discussion focuses on calculating the acceleration of mushroom spores during their launch and subsequent speed reduction. The spore reaches a launch speed of 1.70 m/s over a distance of 5.10 μm, resulting in an acceleration of 2.9 x 104 g during launch. For speed reduction, the spore decelerates to zero over a distance of 1.00 mm, yielding an incorrect initial calculation of 147.45 g, which was later corrected by accounting for the negative acceleration due to deceleration.

PREREQUISITES
  • Understanding of kinematic equations, specifically vf2 - vi2 = 2a distance
  • Basic knowledge of acceleration and deceleration concepts
  • Familiarity with unit conversions, particularly between micrometers and meters
  • Concept of gravitational acceleration (g = 9.8 m/s2)
NEXT STEPS
  • Review kinematic equations for motion under constant acceleration
  • Study the effects of air resistance on projectile motion
  • Explore the mechanics of spore dispersal in fungi
  • Learn about the physics of small-scale movements in biological systems
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Students studying physics, particularly in the context of biological mechanisms, and anyone interested in the dynamics of spore dispersal in fungi.

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Homework Statement


Catapulting mushrooms. Certain mushrooms launch their spores by a catapult mechanism. As water condenses from the air onto a spore that is attached to the mushroom, a drop grows on one side of the spore and a film grows on the other side. The spore is bent over by the drop's weight, but when the film reaches the drop, the drop's water suddenly spreads into the film and the spore springs upward so rapidly that it is slung off into the air. Typically, the spore reaches a speed of 1.70 m/s in a 5.10 μm launch; its speed is then reduced to zero in 1.00 mm by the air. Using that data and assuming constant accelerations, find the acceleration in terms of g during (a) the launch and (b) the speed reduction.


Homework Equations


vf^2 - vi^2 = 2 a distance
so
a = vf^2 / (2 distance)



The Attempt at a Solution


i got part a correct by doing (1.7^2)/2(5.1x10^-6) then dividing by 9.8 to get 2.9x10^4
so i thought part b for speed reduction could use same formula and velocity (1.7) but different distance now (.001m) so i did (1.7^2)/2(.001m)= 1445 then divided by 9.8 for g unit and got 147.45 and the answer is wrong! I need help!
 
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