Time and time intervals when using kinematic equations

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Homework Help Overview

The discussion revolves around a kinematics problem involving a space vehicle that accelerates uniformly. The original poster presents a scenario where the vehicle's initial and final velocities are given, along with a specific time interval, and seeks to understand the displacement during a portion of that interval.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate displacement using different kinematic equations and expresses confusion about the appropriate use of time intervals in these equations. They question the meaning of Δx and the distinction between using t and Δt in their calculations.

Discussion Status

Participants are actively engaging with the original poster's questions, offering insights and prompting further exploration of the equations used. There is a recognition of the need to clarify the initial velocity at specific times, and some participants suggest writing out expressions to analyze the problem more deeply.

Contextual Notes

The original poster indicates a misunderstanding regarding the application of kinematic equations, particularly in relation to initial velocity and time intervals. There is an acknowledgment of the complexity of the problem and the need for careful consideration of the variables involved.

nerf225
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1. A space vehicle accelerates uniformly from 85 m/s at t=0s to 462 m/s at t=10s. What is avg acceleration? How far did the space vehicle move from t=2s to t=6s.

2. Got the avg accel easily enough w basic equation for accel. Problem is conceptual with the displacement portion of the question.

3. I got the correct answer by using x=xo + vit + 1/2at2 for t=2s and t=6s and then basically did Δx= x6s-x2s.

BUT...I don't understand why I can't just use Δx=vit + 1/2at2 with t=4 because the time difference between 6s and 2s is 4s. Isn't the t in that equation essentially a Δt? Shouldn't putting 4 in there tell me the distance it will travel in 4 s?

And why do I get a third different answer if I use: Δx=1/2(vi+vf)Δt?


I seem to be confused about what Δx means and when t vs. Δt is used. Or something. Ugh. Sorry. Help!
 
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Can you tell me what are your results?
 
nerf225 said:
BUT...I don't understand why I can't just use ##\Delta x = v_i t+\frac 12 at^2## with t=4 because the time difference between 6s and 2s is 4s. Isn't the t in that equation essentially a Δt? Shouldn't putting 4 in there tell me the distance it will travel in 4 s?
It will, but you need to plug in the right value for vi. What value did you use?
 
Draw the graph and show us how you did it.
 
nerf225 said:
3. I got the correct answer by using x=xo + vit + 1/2at2 for t=2s and t=6s and then basically did Δx= x6s-x2s.

BUT...I don't understand why I can't just use Δx=vit + 1/2at2 with t=4 because the time difference between 6s and 2s is 4s. Isn't the t in that equation essentially a Δt? Shouldn't putting 4 in there tell me the distance it will travel in 4 s?


You say you got the correct result for the distance covered between those times by taking the difference of the results. Using symbols for the two times, say ##t_a## and ##t_b## for the 2s and 6s times, write out the two expressions for the total distance using the formula. Take the difference of the two expressions and collect the terms. Can you spot the problem with just plugging in ##Δt = t_b - t_a## into your formula?
 
Haruspex -- Thank you! THAT was it! I was plugging in 85 m/s as the initial velocity, not calculating the initial velocity at 2 s and plugging that in. UGH! Smh. Thank you everyone for commenting and being willing to try to help! :)
 

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