Average velocity of a falling object problem

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SUMMARY

The average velocity of a falling object described by the equation x = 100 m - (4.9 s-2)t2 can be calculated using the formula for average velocity, which is Δx/Δt. For the time interval from t = 0 to t = 2 seconds, the initial position is 100 m and the final position is 19.6 m. The average velocity is determined to be 40.2 m/s, calculated as (19.6 m - 100 m) / (2 s - 0 s). The key takeaway is to ensure the time interval is correctly applied in the average velocity formula.

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  • Understanding of kinematic equations, specifically for falling objects.
  • Familiarity with the concept of average velocity.
  • Basic algebra skills for solving equations.
  • Knowledge of units of measurement (meters, seconds).
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  • Learn about the implications of gravitational acceleration on falling objects.
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Violagirl
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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.
 
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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:
 

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