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A particle has displacement equation

(i) xA = 2t + 7

(ii) xB = 3t2 + 2t + 6

(iii) xc = 5t3 + 4t

which of them has uniform acceleration?

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Best answer

(i) v = \(\frac{dx_A}{dt}\,=\,2\)

and a = \(\frac{d^2x_A}{dt^2}\) = 0

as per this eqn. the particle has no acceleration at all.

(ii) v = \(\frac{dx_B}{dt}\)

= 3 × 2t + 2 + 0

= 6t + 2

a = \(\frac{d^2 x_B}{dt^2}\)

Here, acceleration is uniform.

(iii) v = \(\frac{dx_c}{dt}\)

= 5 × 3t2 + 4

= 15t2 + 4

a = \(\frac{dv}{dt}=\frac{d_2x_c}{dt^2}\)

= 15 × 2t = 30t

Here, acceleration depend upon time so it is not uniform.

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