Question

11. A particle moves a distance x in time t according
to equation x = (t + 5)-1. The acceleration of particle
is proportional to [AIPMT (Prelims)-2010]
(1) (Velocity)^3/2 (2) (Distance)^2
(3) (Distance)^-2 (4) (Velocity)^2/3
​

Answer 1

Given distance x=(t+5)−1

velcoity (v) = dt/dx =−(t+5) −2

acceleration (a) = dt/dv=−2(t+5) −3

v 2/3 =−(t+5) −3

acceleration a α v 2/3

Answer 2

Brother your question is wrong it is (t+5)^-1

so now we need acceleration , so by differentiation

a=d^2x/dt2

a=d/dt×d(t+5)^-1/Dt

a=d/dt×-1(t+5)^-2now differentiate it one more time..

a=-2(t+5)^-3

V^3/2=-(t+5)^-3

so A is directly proportional to V^3/2. ....

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