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

Let's see what to do buddy...

Step-by-step explanation:

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`a = \frac{60}{ ({ (1)/(5)) }^(2) } = (60)/( (1)/(25) ) = 60 * 25 \\ \\ a = 15 * 4 * 25 = 15 * 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 = (1)/(2) \: a \: {t}^(2) + v(0) \: t \\`

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`∆x = (1)/(2) * 1500 * 1 + 0 * 1 \\ \\ ∆x = 750 + 0 = 750`

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And we're done.

Thanks for watching buddy good luck.

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