Speed/Dist Question: Homework Solutions

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SUMMARY

The discussion focuses on a physics problem involving the motion of a bag of candy released from a descending helicopter. The bag's speed after 3 seconds is calculated to be -30.4 m/s, indicating downward motion. The distance fallen by the bag must be calculated to determine its position relative to the helicopter, which is also moving downward at 1.0 m/s. The correct approach involves using kinematic equations to find both the speed and distance accurately.

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Homework Statement


Santa arrives at the local mall by helicopter and, as the helicopter steadily descends at 1.0 m/s, he releases a bag of candy to the children waiting below. After 3 seconds:

a. what is the speed of the bag of candy?
b. how far is it below Santa’s helicopter?
c. What are the answers for parts (a) and (b) if Santa’s helicopter is departing, rising steadily at 1.0 m/s?


Homework Equations


For a. v=v0+gt

negative velocity and grav due to descend.

The Attempt at a Solution



a. -1 m/s+(-9.8m/s^2)(3s)
-1+(-29.4)
=-30.4 m/s^2

I need some assistance for the next step...
 
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Do you mean b for the next step? That looks right for the speed of the bag, but check your units. ##\frac{m}{s^{2}}## is a unit of acceleration, not speed. For b you want to calculate the distance the bag falls in that 3 seconds, and then compare that to the location of the helicopter also at 3 seconds.
 

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