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

## the acceleration of the particle is $K \left({v}_{0} + K x\right)$ the particle takes a time $\left(\frac{1}{k}\right) \cdot \ln \left(\frac{v}{v} _ 0\right)$ to attain speed of v velocity varies linearly with displacement with slope of velocity-displacement curve equal to K. all of the above ~correct option given is 4(help me)

Above three options are correct

#### Explanation:

Given the velocity of particle as a function of displacement $x$ as

$v = {v}_{0} + K x$

(1) The acceleration $a$ is given as follows

$a = \setminus \frac{\mathrm{dv}}{\mathrm{dt}} = \setminus \frac{\mathrm{dv}}{\mathrm{dx}} \setminus \frac{\mathrm{dx}}{\mathrm{dt}} = \setminus \frac{\mathrm{dv}}{\mathrm{dx}} v = v \setminus \frac{\mathrm{dv}}{\mathrm{dx}}$

$\setminus \therefore a = \left({v}_{0} + K x\right) \setminus \frac{d}{\mathrm{dx}} \left({v}_{0} + K x\right)$

$a = \left({v}_{0} + K x\right) K$

$= K \left({v}_{0} + K x\right)$

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

$v = \setminus \frac{\mathrm{dx}}{\mathrm{dt}}$

$\mathrm{dt} = \setminus \frac{\mathrm{dx}}{v}$

$\setminus \therefore \mathrm{dt} = \setminus \frac{\mathrm{dx}}{{v}_{0} + K x}$

$\setminus \int \mathrm{dt} = \setminus \int \setminus \frac{\mathrm{dx}}{{v}_{0} + K x}$

$\setminus {\int}_{0}^{t} \mathrm{dt} = \setminus {\int}_{0}^{x} \setminus \frac{\mathrm{dx}}{{v}_{0} + K x}$

$t = {\left[\setminus \frac{\setminus \ln \left({v}_{0} + K x\right)}{K}\right]}_{0}^{x}$

$t = \frac{1}{K} \setminus \ln \left(\setminus \frac{{v}_{0} + K x}{{v}_{0}}\right)$

$t = \frac{1}{K} \setminus \ln \left(\setminus \frac{v}{{v}_{0}}\right) \setminus \quad \left(\setminus \because {v}_{0} + K x = v\right)$

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

$\setminus \frac{\mathrm{dv}}{\mathrm{dx}} = \setminus \frac{d}{\mathrm{dx}} \left({v}_{0} + K x\right) = K$