Time and time intervals when using kinematic equations

AI Thread Summary
A space vehicle accelerates uniformly from 85 m/s to 462 m/s over 10 seconds, prompting a discussion on calculating average acceleration and displacement between specific time intervals. The correct displacement was found using the formula x = x0 + vit + 1/2at² for the times 2s and 6s, but confusion arose regarding the use of Δx = vit + 1/2at² with a Δt of 4 seconds. The key issue was the initial velocity value used; the correct initial velocity at 2 seconds needed to be calculated instead of using the starting velocity of 85 m/s. Participants emphasized the importance of understanding how to apply kinematic equations correctly in relation to time intervals. Clarifying these concepts helped resolve the confusion around displacement calculations.
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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