# I'm having a comprehension issue with a 2-D motion problem.

• LugubriuousLamia
In summary: I'm not sure if it's because the acceleration is constantly changing or what, but it seems like the final velocity would be more accurate if the average speed was taken over a longer period of time.
LugubriuousLamia
1. Problem Statement
A 2 kilogram block rests at the edge of a platform that is 10 meters above level ground.
The block is launched horizontally from the edge of the platform with an initial speed of
3 meters per second. Air resistance is negligible. The time it will take for the block to
reach the ground is most nearly

## Homework Equations

The kinematic equations, specifically to find the final velocity
Vf2= Vi2+2aΔX (to find the final velocity)
t=(Vf-Vi)/2

## The Attempt at a Solution

I believe that the motion in the x direction is irrelevant. I also know that the acceleration is roughly 9.8 (m/s)/s in the direction toward the ground. So I know that the time it should take to go 10 meters is 1 second. However my calculations produce a final velocity of 14 m/s. Thus the time it takes is 1.4 seconds as the final and initial velocities summed is 14 m/s and that divided by 9.8 (m/s)/s is 1.42 seconds. I was able to produce this result knowing that the initial velocity in the Y direction is 0 m/s and that the acceleration and distance are 9.8 (m/s)/s and 10 m respectively. This calculation provides a result of 14 m/s as the final velocity and the time as 1.42 seconds.
This means that the velocity of the object increased by 4.2 m/s over .2 m. I made this conclusion based on the fact that in 1 second the object will have a velocity of 9.8 m/s. So it seems unrealistic that the object that took 1 second and 9.8 meters to accelerate to 9.8 m/s would suddenly accelerate to a velocity of 14 m/s over the course of .2 meters.

I am not sure what I am doing wrong, I believe I may be incorrectly applying the kinematic equations. I know the kinematic equations can be used when the velocity or acceleration is constant which it is here in the y component. I know this because the force of gravity is the only force acting on the object as it's in free-fall. The only error that I can see myself having is from the square root of the 2aΔX (because the acceleration value is negative thus providing a non-real answer). But I believe I could report the acceleration as a scalar descriptive quantity here. That being said I could very well be wrong.

I would appreciate some help. Thank you all very much. Also I apologize if there are any glaring errors or grammatical errors.

-John

Last edited by a moderator:
Why don't you trust your calculations? The total time and final vertical component of velocity look right.

I don't understand your argument about The final 0.2m. Something has gone wrong there perhaps.

PeroK said:
Why don't you trust your calculations? The total time and final vertical component of velocity look right.

I don't understand your argument about The final 0.2m. Something has gone wrong there perhaps.
Thank you for replying, I appreciate it. And I don't trust them because it seems unlikely to me that over a span of 10 meters, with an acceleration value of 9.8 (m/s)/s that the final velocity would be 14 m/s. Which would then produce a time of 1.4 seconds. This means that the time it take for the object to travel 9.8 meters is equal to 1 second and the time it takes to go 10 meters is 1.42 seconds. Thus to travel that final .2 meters is .42 seconds. It seems logically disproportionate.

John Mantia said:
Thank you for replying, I appreciate it. And I don't trust them because it seems unlikely to me that over a span of 10 meters, with an acceleration value of 9.8 (m/s)/s that the final velocity would be 14 m/s. Which would then produce a time of 1.4 seconds. This means that the time it take for the object to travel 9.8 meters is equal to 1 second and the time it takes to go 10 meters is 1.42 seconds. Thus to travel that final .2 meters is .42 seconds. It seems logically disproportionate.

There is an average speed trick, for constant acceleration, that I use but no one seems to teach.

After 1.4s the final speed will be 14m/s, so the average speed is 7m/s. And 7m/s for 1.4s is 9.8m, which is close enough to 10m. That's a sanity check for you.

The problem is that it doesn't take 1s to travel 9.8m. The distance traveled in 1s, using the average speed calculation, is about 5m.

And 5m after 1s and 10m after 1.4s under gravitational acceleration from rest looks about right to me.

Ps try putting the ##s = \frac12 at^2## formula in a spreadsheet and see how far it falls in each successive time interval. That might be quite instructive.

PeroK said:
Ps try putting the ##s = \frac12 at^2## formula in a spreadsheet and see how far it falls in each successive time interval. That might be quite instructive.
I tried to do this but I unfortunately can not figure out how to get the formula to work in excel. By this I mean I have the formula in excel but I can't quite graph the function. I am beginning to grasp your response but still my brain is stuck.

My understanding is that the gravitational constant states that for every second the velocity changes at a rate of 9.8 m/s. After one second of accelerating the velocity should be 9.81 m/s. then reading into that statement the distance traveled should be 9.81m. My issue is that over the remaining ~.2 meters the velocity seems to accelerate from 9.81 m/s to 14 m/s in the duration of .2 meters. The final velocity at 2 seconds should be 19.62 m/s. It just seems that based on my calculations there is an undue acceleration of this mass between 9.81m and 10m. It's one of two things, my application of the math is incorrect or my perception of the relationship between acceleration, velocity and displacement is incorrect. I wish I could figure out the graphing aspect of this because it would bring all of this to light.

John Mantia said:
I tried to do this but I unfortunately can not figure out how to get the formula to work in excel. By this I mean I have the formula in excel but I can't quite graph the function. I am beginning to grasp your response but still my brain is stuck.

My understanding is that the gravitational constant states that for every second the velocity changes at a rate of 9.8 m/s. After one second of accelerating the velocity should be 9.81 m/s. then reading into that statement the distance traveled should be 9.81m. My issue is that over the remaining ~.2 meters the velocity seems to accelerate from 9.81 m/s to 14 m/s in the duration of .2 meters. The final velocity at 2 seconds should be 19.62 m/s. It just seems that based on my calculations there is an undue acceleration of this mass between 9.81m and 10m. It's one of two things, my application of the math is incorrect or my perception of the relationship between acceleration, velocity and displacement is incorrect. I wish I could figure out the graphing aspect of this because it would bring all of this to light.

EDIT: I figured out how to do this and it works graphically but I am not entirely convinced because the math could be correctly proving the wrong thing. I feel that there is something I am missing, however I could just be incorrectly perceiving this.

EDIT: It finally clicked. Because it is accelerating the final velocity after one second is 9.81 m/s but it takes time to get to that speed. Thus it does not travel 9.81 meters in this time. I feel dumb but I am happy I got my answer and it makes sense to me. Thank you for bearing with me as it took me a minute to comprehend what was happening.

If something moves at 9.8m/s for 1s them it moves 9.8m.

But, if something accelerates from rest to 9.8m/s in 1s, it only moves 4.9m.

That's the difference. The average speed, not final speed, is the key for distance moved.

## 1. What is a 2-D motion problem?

A 2-D motion problem involves an object moving in two dimensions, typically represented as the x and y axes. This means the object is moving horizontally and vertically at the same time.

## 2. What causes a comprehension issue with a 2-D motion problem?

There are several factors that can lead to a comprehension issue with a 2-D motion problem. Some common reasons include not understanding the basic principles of motion, not being familiar with the equations and formulas used, or not having enough practice solving similar problems.

## 3. How can I improve my comprehension of 2-D motion problems?

To improve your understanding of 2-D motion problems, it is important to first review the basic principles of motion, such as velocity, acceleration, and displacement. Then, familiarize yourself with the equations and formulas used to solve these types of problems. Practice solving different types of 2-D motion problems to build your skills and confidence.

## 4. Are there any tips for solving 2-D motion problems?

Yes, there are some helpful tips for solving 2-D motion problems. First, draw a clear diagram to represent the given situation. Then, break down the problem into smaller components and use the appropriate equations to solve for each component separately. Finally, double check your calculations and make sure your answers make sense in the context of the problem.

## 5. What are some real-life applications of 2-D motion problems?

2-D motion problems have many real-life applications, including predicting the trajectory of a projectile, analyzing the motion of objects in sports or games, and understanding the motion of objects in a roller coaster or amusement park ride. They are also used in fields such as engineering, physics, and astronomy to study the motion of various objects and systems.

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