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The position x of a particle with respect to time to along x-axis is given by x=9t^2 -t^3 where x in meters and the in sec. What will be the position of this particle when it achieves maximum speed along the positive x-direction?
....
(a) 54
(b) 81
(c) 24
(d) 32?
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Solution:

To find the maximum speed, we need to differentiate the given equation with respect to time t and equate it to zero.

Given, x = 9t^2 - t^3

Differentiating both sides with respect to t, we get

dx/dt = 18t - 3t^2

At maximum speed, dx/dt = 0

Therefore, 18t - 3t^2 = 0

3t(6 - t) = 0

t = 0 or t = 6

Since we are interested in positive x-direction, we consider t = 6.

The position of the particle at t = 6 is given by x = 9(6)^2 - (6)^3 = 54 meters.

Therefore, the correct option is (a) 54.
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The position x of a particle with respect to time to along x-axis is given by x=9t^2 -t^3 where x in meters and the in sec. What will be the position of this particle when it achieves maximum speed along the positive x-direction?.... (a) 54(b) 81(c) 24(d) 32?
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