Question

A body start from rest are moves with uniform acceleration of 60m/1/5^2.what distance does it cover in third second?

Answer 1

Answer:

Let's see what to do buddy...

Explanation:

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$a = \frac{60}{ ({ \frac{1}{5}) }^{2} } = \frac{60}{ \frac{1}{25} } = 60 \times 25 \\ \\ a = 15 \times 4 \times 25 = 15 \times 100 \\ a = 1500 \: \frac{m}{ {s}^{2} }$

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The body start from rest which means :

$v(0) = 0$

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In third second means :

t = 2 -----¢ t = 3 -----¢ ∆t = 1

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We have this equation to find the distance.

$∆x = \frac{1}{2} \: a \: {t}^{2} + v(0) \: t$

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$∆x = \frac{1}{2} \times 1500 \times 1 + 0 \times 1 \\ \\ ∆x = 750 + 0 = 750$

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And we're done.

Thanks for watching buddy good luck.

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