The velocity of particle moving along a straight line increases according to the lawv=v_0+Kxv=v0+Kx , where K is a positive constant,then options are below?

  1. the acceleration of the particle is K(v_0+Kx)K(v0+Kx)
  2. the particle takes a time (1/k)*ln(v/v_0)(1k)ln(vv0) to attain speed of v
  3. velocity varies linearly with displacement with slope of velocity-displacement curve equal to K.
  4. all of the above
    ~correct option given is 4(help me)

1 Answer

Above three options are correct

Explanation:

Given the velocity of particle as a function of displacement xx as

v=v_0+Kxv=v0+Kx

(1) The acceleration aa is given as follows

a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{dv}{dx}v=v\frac{dv}{dx}a=dvdt=dvdxdxdt=dvdxv=vdvdx

\therefore a=(v_0+Kx)\frac{d}{dx}(v_0+Kx)

a=(v_0+Kx)K

=K(v_0+Kx)

(2). We know that the velocity v is given as

v=\frac{dx}{dt}

dt=\frac{dx}{v}

\therefore dt=\frac{dx}{v_0+Kx}

\int dt=\int \frac{dx}{v_0+Kx}

\int_0^t dt=\int_{0}^x \frac{dx}{v_0+Kx}

t=[\frac{\ln(v_0+Kx)}{K}]_{0}^x

t=1/K\ln(\frac{v_0+Kx}{v_0})

t=1/K\ln(\frac{v}{v_0})\quad (\because v_0+Kx=v)

(3) Slope of velocity v vs displacement x curve is given as

\frac{dv}{dx}=\frac{d}{dx}(v_0+Kx)=K