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Reika [66]
3 years ago
15

Please someone derive 3rd equation of motion u- v= 2sa​

Physics
2 answers:
zlopas [31]3 years ago
6 0

We know

\\ \sf\longmapsto a=\dfrac{dv}{dt}

\\ \sf\longmapsto a=\dfrac{dv}{dx}.\dfrac{dx}{dt}

\\ \sf\longmapsto a=v\dfrac{dv}{dx}

\\ \sf\longmapsto adx=vdv

  • Integrate

\\ \sf\longmapsto a{\displaystyle{\int}^x_{x_0}}dx=\displaystyle{\int}_u^v vdv

\\ \sf\longmapsto a(x-x_0)=\dfrac{v^2-u^2}{2}

Here

  • x-x_0=s

\\ \sf\longmapsto v^2-u^2=2as

Kitty [74]3 years ago
3 0

mark me as brainliest ❤️

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pantera1 [17]

(8) A car starting with a speed <em>v</em> skids to a stop over a distance <em>d</em>, which means the brakes apply an acceleration <em>a</em> such that

0² - <em>v</em>² = 2 <em>a</em> <em>d</em> → <em>a</em> = - <em>v</em>² / (2<em>d</em>)

Then the car comes to rest over a distance of

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Doubling the starting speed gives

- (2<em>v</em>)² / (2<em>a</em>) = - 4<em>v</em>² / (2<em>a</em>) = 4<em>d</em>

so the distance traveled is quadrupled, and it would move a distance of 4 • 15 m = 60 m.

Alternatively, you can explicitly solve for the acceleration, then for the distance:

A car starting at 50 km/h ≈ 13.9 m/s skids to a stop in 15 m, so locked brakes apply an acceleration <em>a</em> such that

0² - (13.9 m/s)² = 2 <em>a</em> (15 m) → <em>a</em> ≈ -6.43 m/s²

So the same car starting at 100 km/h ≈ 27.8 m/s skids to stop over a distance <em>d</em> such that

0² - (27.8 m/s)² = 2 (-6.43 m/s²) <em>d</em> → <em>d</em> ≈ 60 m

(9) Pushing the lever down 1.2 m with a force of 50 N amounts to doing (1.2 m) (50 N) = 60 J of work. So the load on the other end receives 60 J of potential energy. If the acceleration due to gravity is taken to be approximately 10 m/s², then the load has a mass <em>m</em> such that

60 J = <em>m g h</em>

where <em>g</em> = 10 m/s² and <em>h</em> is the height it is lifted, 1.2 m. Solving for <em>m</em> gives

<em>m</em> = (60 J) / ((10 m/s²) (1.2 m)) = 5 kg

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