Calculus Problem: Solving for t Given x=6.1

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The problem involves calculating the time t when x=6.1, leading to the equation x(t)=(0.8835+0.05355 t) t^2, which results in t=2.45175. However, using average acceleration to calculate distance with the formula x=1/2at^2 does not yield the expected result of 6.1. The discussion highlights that average acceleration is not suitable for this scenario since acceleration is a function of time. The formula x=1/2at^2 is only valid under constant acceleration conditions. Therefore, the initial solution for t is correct, but the method for calculating distance using average acceleration is flawed.
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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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