Object Distance Homework: Find Distance in 2nd Sec

  • Thread starter josh84
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In summary, the conversation discussed finding the distance an object will fall during the first and second seconds of time after being released from rest. The equations for uniform acceleration were suggested as a way to solve this problem.
  • #1
josh84
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Homework Statement


An object is released from rest and falls a distance H during the first second of time. How far will it fall during the next second of time?


Homework Equations


That's what I need to know.


The Attempt at a Solution



I think it is either 2H or H^2. But I'm not sure.
 
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  • #2
josh84 said:

Homework Statement


An object is released from rest and falls a distance H during the first second of time. How far will it fall during the next second of time?


Homework Equations


That's what I need to know.


The Attempt at a Solution



I think it is either 2H or H^2. But I'm not sure.

again use those three uniform acceleration equations.
Find vf after 1 second
 
  • #3


I would approach this problem by first analyzing the given information and identifying any relevant equations or principles that can be applied. In this case, we are given the initial distance (H) and the time interval (1 second). Using the equation d = 1/2at^2, where d is the distance, a is the acceleration due to gravity (9.8 m/s^2), and t is the time, we can calculate the distance fallen during the first second as H = 1/2(9.8)(1)^2 = 4.9 meters.

To find the distance fallen during the next second, we can use the same equation but with a new time interval of 2 seconds. Plugging in the values, we get d = 1/2(9.8)(2)^2 = 19.6 meters. Therefore, the object will fall a total distance of 19.6 meters during the first 2 seconds.

It is important to note that the distance fallen during the second second is not simply 2H or H^2. These values do not take into account the effects of acceleration due to gravity and only consider the initial distance. It is crucial to use the appropriate equation and units when solving scientific problems.
 

1. How do I calculate the object distance in the 2nd sec?

To calculate the object distance in the 2nd sec, you will need to know the initial velocity, acceleration, and time. You can use the formula: d = ut + 1/2at^2, where d is the distance, u is the initial velocity, a is the acceleration, and t is the time.

2. What is the significance of finding the distance in the 2nd sec?

The 2nd sec represents the second second of motion, which is an important time interval for calculating distance. By finding the distance in the 2nd sec, you can accurately measure the position of an object at a specific time during its motion.

3. Can I use the same formula for any type of motion?

Yes, the formula d = ut + 1/2at^2 can be used for any type of motion, as long as you have the initial velocity, acceleration, and time values. However, it is important to note that this formula assumes constant acceleration, so it may not be accurate for objects with varying acceleration.

4. How do I measure the initial velocity and acceleration for an object?

The initial velocity can be measured by recording the starting position and time of the object, and then using the formula v = d/t to calculate its velocity. The acceleration can be measured by recording the change in velocity over time, using the formula a = (v-u)/t.

5. Is there a specific unit for the distance calculated in the 2nd sec?

The distance calculated in the 2nd sec can be measured in any unit of length, such as meters, feet, or kilometers. It is important to be consistent with units when using the formula, and to convert to the desired unit if needed.

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