Question

Particle moves a distance $x$ in time $t$ according to equation $x=(t+5)^{-1}$ The acceleration of particle is proportional to

• A. (velocity)$^{3/2}$
• B. (distance)$^2$
• C. (distance)$^{-2}$
• D. (velocity)$^{2/3}$

Answer 1

Answer: (A)

Distance, $x=(t+5)^{-1}\,\,\,\,$....(i)
Velocity, $v=\frac{d x}{d t}=\frac{d}{d t}(t+5)^{-1}$
$=-(t+5)^{-2}\,\,\,\,$....(ii)
Acceleration, $a=\frac{d v}{d t}=\frac{d}{d t}\left[-(t+5)^{-2}\right]$
$=2(t+5)^{-3}\,\,\,\,$...(iii)
From equation (ii), we get
$v^{3 / 2}=-(t+5)^{-3}\,\,\,$...(iv)
Substituting this in equation (iii) we get
Acceleration, $a=-2 v^{3 / 2}$
or $ a \propto(\text { velocity })^{3 / 2}$
From equation (i), we get
$x^{3}=(t+5)^{-3}$
Substituting this in equation (iii), we get
Acceleration, $a=2 x^{3}$
or $ a \propto$ (distance $^{3}$