How Do You Calculate Instantaneous Velocity at Point 2?

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Homework Help Overview

The discussion revolves around calculating the instantaneous velocity at a specific point in a motion scenario involving three points with varying distances and time intervals. The subject area is kinematics, focusing on concepts of velocity and acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between distance and time to determine instantaneous velocity, questioning whether to average velocities from adjacent intervals. There are inquiries about the nature of the motion, specifically regarding constant versus variable acceleration.

Discussion Status

The discussion is active, with participants providing background information and asking clarifying questions. Some guidance is offered regarding the calculation of average acceleration and its relation to instantaneous velocity, but no consensus has been reached on the method to be used.

Contextual Notes

Participants note the distances between points and the time intervals, indicating that the motion involves variable acceleration. There is an emphasis on understanding the definitions of acceleration and velocity in this context.

deficiency4math
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Hi,

I was wondering how to solve the instantaneous velocity for this.

I have 3 points which have 5 cm between point 1 and point 2 and 9 cm between point 2 and point 3 (this is easy, i just need to grasp the concept)

The question is "What is the instantaneous velocity for point 2?

How would I figure this out?

thanks :smile:
 
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Do you have any times?
 
oh, yes, sorry,

the time intervals between the points are 1/40th of a second (lets say)

(sorry, I forgot to post them... I kinda forgot!)
 
Can you give a little background? What kind of motion is going on here, constant acceleration?
 
it would be variable acceleration, because the consecutive distance between 1&2 is 5 and 2&3 9... so let's say that each time, the velocity increases by one.

Would I just take the velocity of 1-2 (5cm per 1/40th of a second) and 2-3 (9cm per 1/40th of a second) and find the average of them to get the instantaneous?
 
deficiency4math said:
it would be variable acceleration, because the consecutive distance between 1&2 is 5 and 2&3 9... so let's say that each time, the velocity increases by one.

Define acceleration in terms of velocity. Does acceleration have to change for velocity to change?

Would I just take the velocity of 1-2 (5cm per 1/40th of a second) and 2-3 (9cm per 1/40th of a second) and find the average of them to get the instantaneous?

I don't know how accurate you need to be, but what I would do is try to figure out the average acceleration. The object is speeding up, so its position function is:

[tex]x(t) = x_0 + v_0t + 1/2 * at^2[/tex]

We traveled x(t) = 14cm in t=0.05sec. Can you find the acceleration? Can you find the velocity at t=0.025 with this?
 

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