Question

A particle is moving along the X-axis. Its accelertaiOn at time t is proportional to its velocity at that time. Find the differential equation of the motion of the particle.

Answer 1

the answer is wxyz ya dig

Answer 2

Answer:

$\dfrac{\mathrm{d}^2 x(t)}{\mathrm{d} t^2}-K\dfrac{\mathrm{d} x(t)}{\mathrm{d} t}=0$

where, $x(t)$ is the position of the particle at time t.

Step-by-step explanation:

Given:

The acceleration of the particle at time t is proportional to its velocity at that time.

We know,

• The acceleration of a particle at a given time is defined as the rate of change of velocity of the particle at that time.

$\vec a(t) = \dfrac{\mathrm{d} \vec v(t)}{\mathrm{d} t}.$

• The velocity of a particle at a given time is defined as the rate of change of position of the particle at that time.

$\vec v(t) = \dfrac{\mathrm{d} \vec r(t)}{\mathrm{d} t}.$

where, $x(t)$ is the position of the particle at time t.

Given that the particle is moving along the x-axis only, therefore, acceleration and the velocity of the particle are only along the x axis direction.

As per the given situation,

$a(t) \propto v(t)$.

The vector sign could be removed because the particle is moving along x direction only.

On putting the values of acceleration and velocity of the particle,

$\dfrac{\mathrm{d} v(t)}{\mathrm{d} t}\propto v(t)\\\dfrac{\mathrm{d} v(t)}{\mathrm{d} t}=K v(t)\ \ \ \ \ \ \ \ K\text{ is the constant of proportionality}\\\dfrac{\mathrm{d} }{\mathrm{d} t}\left (\dfrac{\mathrm{d} x(t)}{\mathrm{d} t} \right )=K\dfrac{\mathrm{d} x(t)}{\mathrm{d} t}\\\dfrac{\mathrm{d}^2 x(t)}{\mathrm{d} t^2}=K\dfrac{\mathrm{d} x(t)}{\mathrm{d} t}\\\dfrac{\mathrm{d}^2 x(t)}{\mathrm{d} t^2}-K\dfrac{\mathrm{d} x(t)}{\mathrm{d} t}=0.$

It is the required differential equation.