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