Average velocity of a falling object problem

Click For Summary
The problem involves calculating the average velocity of a falling object using the height equation x = 100 m - (4.9 s-2)t². The initial height at t = 0 is 100 m, and at t = 2 s, the height is 19.6 m. The average velocity is determined by the formula Δx/Δt, where Δx is the change in height and Δt is the time interval of 2 seconds. The confusion arose regarding the time interval and the cancellation of units, but clarification helped resolve the misunderstanding. The final average velocity can be calculated using the correct values for height and time.
Violagirl
Messages
112
Reaction score
0

Homework Statement



A falling object moves so that its height x above the ground at time t is given by the following equation: x = 100 m - (4.9 s-2)t2. Find its average velocity from t = 0 t t = 2 s.


Homework Equations



Average velocity = Δx/Δt

The Attempt at a Solution



I was able to work it out to the point where I solved for x by plugging in the times given for 0 secs and 2 secs.

For t = 0:

x = 100 m - (4.9 s -2 * (0 s)2

= 100 m

For t = 2:

x = 100 m - (4.9 s -2 * (2 s)2

= 19.6 m

Solving for t2 in the equation, I get:

100 m/4.9 s-2 = t2

Final answer of 4.52 m s -2

For the average velocity equation (and t), I realize this does not work all that great as m will be canceled in setting up the average velocity equation and am only left with seconds for time. I'm not sure what it is that I'm doing wrong in this problem. Thank you everyone for your help.
 
Physics news on Phys.org
Violagirl said:

Homework Statement



A falling object moves so that its height x above the ground at time t is given by the following equation: x = 100 m - (4.9 s-2)t2. Find its average velocity from t = 0 t t = 2 s.


Homework Equations



Average velocity = Δx/Δt

The Attempt at a Solution



I was able to work it out to the point where I solved for x by plugging in the times given for 0 secs and 2 secs.

For t = 0:

x = 100 m - (4.9 s -2 * (0 s)2

= 100 m

For t = 2:

x = 100 m - (4.9 s -2 * (2 s)2

= 19.6 m

Solving for t2 in the equation, I get:

100 m/4.9 s-2 = t2

Final answer of 4.52 m s -2

For the average velocity equation (and t), I realize this does not work all that great as m will be canceled in setting up the average velocity equation and am only left with seconds for time. I'm not sure what it is that I'm doing wrong in this problem. Thank you everyone for your help.

Average velocity is just (final position - initial position) divided by time interval.
You have successfully found the initial and final positions.
You just need the time interval.

How long was it from t = 0 to t = 2 ?
 
PeterO said:
Average velocity is just (final position - initial position) divided by time interval.
You have successfully found the initial and final positions.
You just need the time interval.

How long was it from t = 0 to t = 2 ?

Oh ok, got it! I guess I was just overthinking it, oops :redface:...Thanks a lot! :biggrin:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
5
Views
2K
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
887
Replies
20
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K