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The velocity of a particle moving on the x-axis is given by v = x^2 + x, where x is in m and v is in m/s. What is its position (in m) when its acceleration is 30m/s^2

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4 years ago

```							accln. = vdv/dx = > (x^2 +x)*(2x + 1) = 30 => 2x^3 + x^2 + 2x^2 + x – 30 = 0From Remainder tgeorem x = 2 satifies it.one root is x = 2.Now to find other roots,a + b + 2 = -0.5a + b = – 2.5and 2ab = 15ab = 15/2the eqn. becomes,a + 15/2a = -2.5a^2 + (5/2)a + 15/2 = 0or2a^2 + 5a + 15 = 0It’s Discriminant is less than zero hence it has imaginafry roots.Therefore.the position is at x= 2 meter.
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4 years ago
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