A particle moves along the positive x-direction such that its speed depends on position according to v = $v_{o}$ - kx, where $v_{o}$ and k are positive constants and v remains positive in the region considered. The acceleration of the particle as a function of position x is:
Options

Question

A particle moves along the positive x-direction such that its speed depends on position according to v = $v_{o}$ - kx, where $v_{o}$ and k are positive constants and v remains positive in the region considered. The acceleration of the particle as a function of position x is:

Options

G1 Physics · Accepted Answer

Using a = dv/dt = (dv/dx)(dx/dt) = v(dv/dx). Since v = v0 - kx, dv/dx = -k. Thus a = (v0 - kx)(-k) = k 2 x k^2x k 2 x - kv0. Hence, option B is correct.

A particle moves along the positive x-direction such that its speed depends on position according to v = v o v_{o} v o ​ - kx, where v o v_{o} v o ​ and k are positive constants and v remains positive in the region considered. The acceleration of the particle as a function of position x is:

A particle moves along the positive x-direction such that its speed depends on position according to v = v o v_{o} v o - kx, where v o v_{o} v o and k are positive constants and v remains positive in the region considered. The acceleration of the particle as a function of position x is: