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Grade: 12th pass

                        

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

4 years ago

Answers : (1)

Vikas TU
12278 Points
							
accln. = vdv/dx = > (x^2 +x)*(2x + 1) = 30 
=> 2x^3 + x^2 + 2x^2 + x – 30 = 0
From Remainder tgeorem x = 2 satifies it.
one root is x = 2.
Now to find other roots,
a + b + 2 = -0.5
a + b = – 2.5
and 
2ab = 15
ab = 15/2
the eqn. becomes,
a + 15/2a = -2.5
a^2 + (5/2)a + 15/2 = 0
or
2a^2 + 5a + 15 = 0
It’s Discriminant is less than zero hence it has imaginafry roots.
Therefore.
the position is at x= 2 meter.
4 years ago
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