Given, distance x=(t+5)−1 Differentiating Eq. (i) w.r.t. t, we get dtdx=(v)=(t+5)2−1 Again, differentiating Eq. (i) w.r.t. t, we get dt2d2x=(a)=(t+5)32 Comparing Eqs. (ii) and (iii), we get (a) ∝(v)3/2
A particle moves a distance x in time t according to equation x=(t+5)−1. The acceleration of particle is proportional to
Held on 30 Apr 2010 · Verified 9 Jul 2026.
(velocity) 3/2
( distance )2
(distance )−2
(velocity) 2/3
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