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Help Finding Average Acceleration

  • #1

Homework Statement


NASA operates a 2.2 second drop tower at the Glenn Research Center in Ohio, where experimental packages are dropped from the top of the tower.
a) What is the drop distance of a 2.2 s tower?
b) How fast are the experiments traveling when they hit the airbags at the bottom of the tower?
c) If the experimental package comes to rest over a distance of .75 m upon hitting the airbags, what is the average stopping acceleration?


The Attempt at a Solution


I'm pretty sure I've got a & b correct, but I'm not sure how to go about solving part c. Can someone help me please?

a) What is the drop distance of a 2.2 s tower?
x = 1/2*gt2 = .5*9.81*2.22 = 23.74 m

b) How fast are the experiments traveling when they hit the airbags at the bottom of the tower?
v = gt = 9.81 * 2.2 = 21.582 m/s

c) If the experimental package comes to rest over a distance of .75 m upon hitting the airbags, what is the average stopping acceleration?
I think one formula for acceleration is Δv/Δt but I'm not sure how to use that in this case...
 

Answers and Replies

  • #2
188
1

Homework Statement


NASA operates a 2.2 second drop tower at the Glenn Research Center in Ohio, where experimental packages are dropped from the top of the tower.
a) What is the drop distance of a 2.2 s tower?
b) How fast are the experiments traveling when they hit the airbags at the bottom of the tower?
c) If the experimental package comes to rest over a distance of .75 m upon hitting the airbags, what is the average stopping acceleration?


The Attempt at a Solution


I'm pretty sure I've got a & b correct, but I'm not sure how to go about solving part c. Can someone help me please?

a) What is the drop distance of a 2.2 s tower?
x = 1/2*gt2 = .5*9.81*2.22 = 23.74 m

b) How fast are the experiments traveling when they hit the airbags at the bottom of the tower?
v = gt = 9.81 * 2.2 = 21.582 m/s

c) If the experimental package comes to rest over a distance of .75 m upon hitting the airbags, what is the average stopping acceleration?
I think one formula for acceleration is Δv/Δt but I'm not sure how to use that in this case...
It's worth trying to commit the suvat equations to memory:
s = length, u = initial speed, v = final speed .... a and t you can probably guess!

v = u + at
s = ut + 0.5 at2
s = 0.5(u+v)t
v2 = u2 + 2as
s = vt - 0.5 at2
 
  • #3
16
0

Homework Statement


NASA operates a 2.2 second drop tower at the Glenn Research Center in Ohio, where experimental packages are dropped from the top of the tower.
a) What is the drop distance of a 2.2 s tower?
b) How fast are the experiments traveling when they hit the airbags at the bottom of the tower?
c) If the experimental package comes to rest over a distance of .75 m upon hitting the airbags, what is the average stopping acceleration?


The Attempt at a Solution


I'm pretty sure I've got a & b correct, but I'm not sure how to go about solving part c. Can someone help me please?

a) What is the drop distance of a 2.2 s tower?
x = 1/2*gt2 = .5*9.81*2.22 = 23.74 m

b) How fast are the experiments traveling when they hit the airbags at the bottom of the tower?
v = gt = 9.81 * 2.2 = 21.582 m/s

c) If the experimental package comes to rest over a distance of .75 m upon hitting the airbags, what is the average stopping acceleration?
I think one formula for acceleration is Δv/Δt but I'm not sure how to use that in this case...
You need to determine the stopping time.
 
  • #4
1,065
10
For question (c) its energy problem. Refer to problem(b)
 
  • #5
188
1
You need to determine the stopping time.
For question (c) its energy problem. Refer to problem(b)
The distance, initial and final velocities are known. Does this situation not seem to be simply covered by one of the standard suvat equations?
 
  • #6
16
0
The distance, initial and final velocities are known. Does this situation not seem to be simply covered by one of the standard suvat equations?
Indeed, but the OP seemed to have trouble determining which one.
 

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