Calculating x Position of a Moving Object

In summary: Consider the ##x## component of displacement, velocity, and acceleration for both L to M as well as from M to N .Thank you for your help! I figured it out!The answer is t = 0.572s.
  • #1
Undeterred247
6
4
Homework Statement
A very small steel marble is shown rolling at a constant speed on a horizontal table. The marble leaves the table at M, falls, and hits the ground at N. This is illustrated in the diagram below which is drawn to scale. Calculate the time it took the marble to travel from L to N.
Relevant Equations
p(t) = p0 +v0 * t +1/2*a*t^2
a = -g = -9.81m/s^2 = -981cm/s^2
Vertical position (x component):
y(t) = y0 +v0 * t -1/2*g*t^2
Position yx(t) = 0cm = 40cm + 0 *t -1/2 * 981cm/s/s * t^2
edit: I replaced v0 = 0 and got t = 0.286s, which is incorrect according to the submission.

This didn't work so I thought maybe by finding the angle of the throw would help with the following equation but there are still not enough parts to it:
yx(t) = y0 +v0 * sintheta * t -1/2 *g * t^2
yx(t) = 40cm + v0 * sin(36.84) * t -1/2 *981cm/s/s *t^2

Theta = angle of throw where
horizontal = 30cm
vertical = 40cm
hypotenuse = 50cm

I found https://www.physicsforums.com/threads/rolling-marble-on-table.469634/ which is similar to this problem but I still don't understand how to solve this. Please help!
 

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  • #2
Start with eliminating all aspects of the x-axis.

That leaves a marble that starts off with v=0 and falls a given distance. You should be able to determine the time of that fall.
 
  • #3
Watch this video. What does it suggest about the time of flight of the marble?

 
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  • #4
Undeterred247 said:
Homework Statement:: A very small steel marble is shown rolling at a constant speed on a horizontal table. The marble leaves the table at M, falls, and hits the ground at N. This is illustrated in the diagram below which is drawn to scale. Calculate the time it took the marble to travel from L to N.
Relevant Equations:: p(t) = p0 +v0 * t +1/2*a*t^2
a = -g = -9.81m/s^2 = -981cm/s^2

Vertical position (x component):
y(t) = y0 +v0 * t -1/2*g*t^2
Position yx(t) = 0cm = 40cm + 0 *t -1/2 * 981cm/s/s * t^2
edit: I replaced v0 = 0 and t = 0.286s, which is incorrect according to the submission.

This didn't work so I thought maybe by finding the angle of the throw would help with the following equation but there are still not enough parts to it:
yx(t) = y0 +v0 * sintheta * t -1/2 *g * t^2
yx(t) = 40cm + v0 * sin(36.84) * t -1/2 *981cm/s/s *t^2

Theta = angle of throw where
horizontal = 30cm
vertical = 40cm
hypotenuse = 50cm

I found https://www.physicsforums.com/threads/rolling-marble-on-table.469634/ which is similar to this problem but I still don't understand how to solve this. Please help!
Hello @Undeterred247 .
:welcome:

You correctly calculated the time to go from M to N .

You're not finished.
 
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  • #5
Undeterred247 said:
edit: I replaced v0 = 0 and t = 0.286s, which is incorrect according to the submission.
(I assume you meant "I replaced v0 = 0 and got t = 0.286s".)
Maybe you did not read the question correctly. It says "Calculate the time it took the marble to travel from L to N.", not M to N.
 
  • #6
SammyS said:
Hello @Undeterred247 .
:welcome:

You correctly calculated the time to go from M to N .

You're not finished.
Thank you! How would I use this time from M to N to calculate for the time from L to N?
 
  • #7
Undeterred247 said:
Thank you! How would I use this time from M to N to calculate for the time from L to N?
Do you know how far the ball rolls horizontally from L to N?
Do you know how fast it rolls?
 
  • #8
jbriggs444 said:
Do you know how far the ball rolls horizontally from L to N?
Do you know how fast it rolls?
I calculated
Vy = Voy +ay *t
Vy = 0 -981 * (0.286)^2
Vy = -80.24

Then took the total distance 60cm/-80.24 = t = -0.748
But when checked this answer is incorrect.
 
  • #9
Undeterred247 said:
I calculated
Vy = Voy +ay *t
Vy = 0 -981 * (0.286)^2
Vy = -80.24

Then took the total distance 60cm/-80.24 = t = -0.748
But when checked this answer is incorrect.
Well, yes, that calculation seems to be an incorrect calculation of the wrong thing and does not make any use of the suggestion that was offered.

As nearly as I can determine, it is a garbled calculation of twice the vertical displacement from M to N using cgs units. This is a rather backwards way to proceed since the I believe that the 0.286 s input to the calculation was originally found by using the actual displacement of -40.0 cm as the input.

The factor of two error seems to be from misquoting the SUVAT equation and the 0.24 error is likely from rounding 0.285568... to 0.286. Once that gets fixed, there is the problem of having calculated the wrong thing.
 
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  • #10
Undeterred247 said:
Thank you! How would I use this time from M to N to calculate for the time from L to N?
Consider the ##x## component of displacement, velocity, and acceleration for both L to M as well as
from M to N .
 
  • #11
Thank you for your help! I figured it out!
The answer is t = 0.572s
 
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FAQ: Calculating x Position of a Moving Object

1. How do you calculate the x position of a moving object?

To calculate the x position of a moving object, you need to know the object's initial position, velocity, and time. You can use the formula x = x0 + vt, where x0 is the initial position, v is the velocity, and t is the time.

2. What units should be used for the x position calculation?

The units used for the x position calculation will depend on the units used for the initial position, velocity, and time. It is important to use consistent units, such as meters for distance and seconds for time, to ensure accurate calculations.

3. Can the x position of a moving object change over time?

Yes, the x position of a moving object can change over time if the object is in motion. The position will change as the object moves and its velocity changes. However, if the object is at rest, the x position will remain constant.

4. How does acceleration affect the x position of a moving object?

Acceleration can affect the x position of a moving object by changing its velocity over time. If the object is accelerating, its velocity will increase or decrease, causing its x position to change accordingly.

5. Are there any limitations to using the x position formula for calculating the position of a moving object?

The x position formula can only be used for objects moving in a straight line with constant velocity. It does not take into account other factors such as air resistance or changes in direction. For more complex motion, other formulas or methods may need to be used.

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