Why Is Displacement Only 4.9m After 1 Second in Free Fall?

In summary, the individual displacements at 1 second and 2 seconds are calculated using the formula 1/2*b*h and are found to be 4.9 m and 19.6 m respectively. However, this formula only gives the total displacement from the initial time interval and the actual displacement at those specific times must be calculated by finding the difference between each displacement. The mistake in the red statement is due to not considering the initial velocity of the object.
  • #1
mirza21
3
0
Hi friends am new to Physics. Am facing a problem while caculating displacment.

Acceleration of free falling object is 9.8m/s/s. Am using formula 1/2*b*h to get the displacement.According to me if object acceleration is 9.8 m/s/s then after 1 second it's displacement should be 9.8 meters and after 2 second it should be positioned at 29.4 m.
If i use the formula to calculate displacement then it is positioned at 4.9 m at end of 1st second and a the end of 2nd second it 19.6 m. Please explain why object is positon at 4.9 m by using this formula.


At t = 1 s
d = (0.5) * (9.8 m/s2) * (1 s)2 = 4.9 m

At t = 2 s

d = (0.5) * (9.8 m/s2) * (2 s)2 = 19.6 m

At t = 5 s

d = (0.5) * (10 m/s2) * (3 s)2 = 44.1 m

For refrence see http://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L5d.html
 
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  • #2
mirza21 said:
Hi friends am new to Physics. Am facing a problem while caculating displacment.

Acceleration of free falling object is 9.8m/s/s. Am using formula 1/2*b*h to get the displacement.According to me if object acceleration is 9.8 m/s/s then after 1 second it's displacement should be 9.8 meters and after 2 second it should be positioned at 29.4 m.
If i use the formula to calculate displacement then it is positioned at 4.9 m at end of 1st second and a the end of 2nd second it 19.6 m. Please explain why object is positon at 4.9 m by using this formula.


At t = 1 s
d = (0.5) * (9.8 m/s2) * (1 s)2 = 4.9 m

At t = 2 s

d = (0.5) * (9.8 m/s2) * (2 s)2 = 19.6 m

At t = 5 s

d = (0.5) * (10 m/s2) * (3 s)2 = 44.1 m

For refrence see http://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L5d.html

The statement in red is totally wrong.It also depends on the initial velocity of the object from which it was thrown downwards.
NOW for the hint part:

It seems you are confused with the formula part.
The formula s=1/2 at^2 [u=0] does not give the displacement in that particular time interval,in fact It gives you the total displacement from the initial time interval.
To get the displacement in a particular time interval you must calculate the displacements using the formula and substituting the correct values and then calculate the difference.
 
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  • #3



Hello and welcome to the world of physics! It sounds like you are trying to understand the concept of displacement and how it relates to acceleration. Let's break it down step by step.

Displacement is a measurement of an object's change in position from its starting point to its ending point. It is different from distance, which is the total length traveled by an object. In your case, the displacement is the difference between where the object started and where it ended up.

The formula you are using, d = 1/2 * a * t^2, is actually the formula for calculating the distance an object travels under constant acceleration. This is different from displacement because it takes into account the total distance traveled, not just the change in position.

To calculate displacement, we use the formula d = v0 * t + 1/2 * a * t^2, where v0 is the initial velocity (in this case, 0 since the object is starting from rest). This formula takes into account the object's starting position and its change in position due to acceleration.

Now, let's look at your example. At t = 1 s, the object has an acceleration of 9.8 m/s^2, meaning it is speeding up by 9.8 m/s every second. Using the formula for displacement, we get d = 0 * 1 + 1/2 * 9.8 * 1^2 = 4.9 m. This means that after 1 second, the object has traveled a distance of 4.9 m, but its displacement is only 4.9 m from its starting point.

At t = 2 s, the object has an acceleration of 9.8 m/s^2 and a starting velocity of 0 m/s. Using the displacement formula, we get d = 0 * 2 + 1/2 * 9.8 * 2^2 = 19.6 m. This means that after 2 seconds, the object has traveled a distance of 19.6 m, but its displacement is only 19.6 m from its starting point.

It's important to note that displacement is a vector quantity, meaning it has both magnitude (size) and direction. In your example, the displacement after 1 second is 4.9 m, but it is also in the downward direction due to gravity. This is why it is important to use
 

Related to Why Is Displacement Only 4.9m After 1 Second in Free Fall?

1. What is displacement?

Displacement is a physical quantity that measures the shortest distance between an initial and final position of an object. It is a vector quantity, meaning it has both magnitude and direction.

2. How is displacement different from distance?

Distance is the total length of the path traveled by an object, while displacement is the straight-line distance between the initial and final positions. Distance is a scalar quantity, meaning it only has magnitude.

3. What causes displacement?

Displacement is caused by a change in an object's position or location. This can be due to various factors such as acceleration, force, or motion.

4. How is displacement measured?

Displacement is measured in units of length, such as meters or centimeters. It can be calculated by subtracting the initial position from the final position of an object.

5. What are some real-life examples of displacement?

Examples of displacement can be seen in everyday situations, such as a car driving from one point to another, a person walking from their house to the store, or a pendulum swinging back and forth. Displacement is also used in navigation and mapping, as well as in physics and engineering calculations.

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