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प्रश्न
A particle moves along x-axis obeying the equation x = t (t – 1)(t – 2), where x (in metres) is the position of the particle at any time t (in seconds). The displacement when the velocity of the particle is zero, is
विकल्प
`2/(3sqrt(3)) m, 2/(3sqrt(3)) m`
`5/(3sqrt(3)) m, 5/(3sqrt(3)) m`
– 3 m, 3 m
– 5 m, 5 m
उत्तर
`2/(3sqrt(3)) m, 2/(3sqrt(3)) m`
Explanation:
Displacement x = t(t – 1)(t – 2) ......(i)
∴ x = t3 – 2t2 – t2 + 2t
Velocity v = `(dx)/(dt)`
∴ v = 3t – 6t + 2
For v = 0, we have
3t2 – 6t + 2 = 0 ......(ii)
Equation (ii) is quadratic in t
Solving we get,
t = `(1 + 1/sqrt(3))` or t = `(1 - 1/sqrt(3))` ......(iii)
Now; x = t(t – 1) (t – 2) ......[From (i)]
Substitute equation (iii) in equation (i)
x = `(1 + 1/sqrt(3)) [1 + 1/sqrt(3) - 1] [1 + 1/sqrt(3) - 2]`
Solving we get,
x = `2/(3sqrt(3) m`
Similarly, By substituting the other value of t from equation (iii) in equation (i) and solving we get,
x = `2/(3sqrt(3)) m`