Calculus Problem: Solving for t Given x=6.1

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SUMMARY

The discussion centers on solving for time (t) in the equation x(t) = (0.8835 + 0.05355t)t², given x = 6.1. The user calculated t to be approximately 2.45175. However, they encountered discrepancies when applying the average acceleration formula x = 1/2at², which is only valid under constant acceleration conditions. The conclusion emphasizes that since acceleration is a function of time in this scenario, using average acceleration does not yield the correct distance traveled.

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Homework Statement


Well I've pretty much boiled the problem down to this:

Let a = 1.767+0.3213t
v(0)=0
x(0)=0

then I got
x(t)=(0.8835+0.05355 t) t^2

Given x=6.1

Solving for t I got t=2.45175

But I'm just curious. If I calculate average acceleration I get a_av=2.16.

If I now apply x=1/2at^2 using a average I don't get 6.1

Homework Equations


Using calculus I turned a=1.767+0.3213t to
x=(0.8835+0.05355 t) t^2

Given x is 6.1 I solve t=2.45175


The Attempt at a Solution



But now when I use the average velocity trick and then use x= 1/2at^2 I don't get 6.1. Is my solution incorrect?
 
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Your acceleration is a function of time. The average acceleration won't give you the correct distance traveled.
 
jeremy5561 said:
If I now apply x=1/2at^2 using a average I don't get 6.1
That formula is only valid for constant acceleration.
 

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