A particle moves a distance x in time t according to equation x = (t + 5)-1. The acceleration of particle is proportional to

Given, x=1t + 5Differentiating both sides w.r.t. t, we get            dxdt=-1(t+5)2×1∴  Velocity, v=dxdt=-1(t+5)2                                         …(i)Again, differentiating both sides w.r.t. t, we get            dvdt=1×2(t+5)3(1)=2(t+5)3                                      …(ii)Now, from Eqs. (i) and (ii), we get            dvdt=-2v×1(t+5)=(-2v)(v)=-2(v)3/2∴ Acceleration, a=dvdt=-2v3/2⇒a∝ (velocity) 3/2