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2 years ago

8

An object travels a distance d with acceleration a over a period of time t according to the equation: d = at² After 2.3 seconds

with an acceleration of 9.8 m/s2, how far does an object travel (rounding the answer to two significant figures)?

1 answer:

matrenka [14]2 years ago

5 0

Answer:

A 26m

Explanation:

firstly the answer will not be part c as distance unit will be meters (m)

next, substitute the values of a and t as shown

$\frac{1}{2} (9.8)(2.3)^2 = 25.921$

the answer rounded off to 2 significant figures will be hence 26m

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Answer:

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Explanation:

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or

$\frac{dp}{dI}=\frac{(I^2+I+9).90-90I.(2I+1)}{(I^2+I+9)^2}$

or

$\frac{dp}{dI}=\frac{90I^2+90I+810)-(180I^2+90I)}{(I^2+I+9)^2}$

or

$\frac{dp}{dI}=\frac{-90I^2+810)}{(I^2+I+9)^2}$

Now for the maxima $\frac{dP}{dI}=0$

thus,

$0=\frac{-90I^2+810)}{(I^2+I+9)^2}$

or

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or

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or

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or

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Explanation:

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where,

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