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A particle moves in a line so its position at any time in t ≥0 is given by the function s(t)=t²-6t+5,
where s is measured in meres and t is measured in seconds.
a) find the displacement during the first 6 seconds.
All i did was input 0 into the equation solve, then do the same with 6 and subtract the two. I got 0. Is this the right way to do that, os is that too "physics like"?
b)Find avg. velocity during first 6 seconds.
i basically did the same thing with (y2-y1)/(x2-x1) and got 0.
c) Find instantaneous velocity when t=4.
the derivative of the function is 2t-6 so:
s(4)=2(4)-6
=2
d) Find the acceleration of the particle when t=4
the derivative of 2t-6 is 2 so is it 2?
e) At what values does the particle change direction?
I got when t=3 but only by looking at the graph. How do you do this by calculus?
f) Where is the particle when s is a minimum?
again by using the graph
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I agree with your methods for a. to d. The way you did a. isn't 'too physics-like', it's right.
You possibly found e. difficult because it's actually very simple, and perhaps you missed the method. The particle changes direction simply when the velocity changes sign, or in other words, when it is at 0. You already found the velocity in c. to be 2t-6, and this is 0 when t=3.
f. follows on from e. The displacement is at a minimum when its derivative is 0. And in e, you just found that the velocity was 0 at t=3. So there's your answer. The displacement at that point would be -4.
Why did the vector cross the road?
It wanted to be normal.
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