Question

The acceleration of a particle is defined by the relation a 5 2kv2.5, where k is a constant. The particle starts at x 5 0 with a velocity of 16 mm/s, and when x 5 6 mm, the velocity is observed to be 4 mm/s. Determine (a) the velocity of the particle when x 5

Answer 1

Answer:

The velocity of the particle is 4.76 mm/s.

Explanation:

Given that,

At x = 0, v = 16 mm/s

At x = 6 mm, v = 4 mm/s

The equation of acceleration is

$a=-kv^{2.5}$

Here, k = constant

v= velocity

Th velocity of the particle along straight line

$v=\dfrac{dx}{dt}$

$dt=\dfrac{dx}{v}$....(I)

The acceleration of the particle along a straight line

$a=\dfrac{dv}{dt}$

$dt=\dfrac{dv}{a}$....(II)

From equation (I) and (II)

$\dfrac{dx}{v}=\dfrac{dv}{a}$

$a=v\dfrac{dv}{dx}$

Put the value of a

$-kv^{2.5}=v\dfrac{dv}{dx}$

$\dfrac{dv}{dx}=-kv^{1.5}$

On integrating

$\dfrac{2}{\sqrt{v}}=-kx+C$....(III)

Put the value of x = 0 and v = 16 mm/s in equation (III)

$\dfrac{2}{\sqrt{16}}=0+C$

$C=0.5$

Now put the value of x = 6mm and v = 4 mm in the equation (III)

$\dfrac{2}{\sqrt{4}}=-k\times6+0.5$

$-k\times6=1-\dfrac{1}{2}$

$k=-\dfrac{1}{12}$

Put the value of k and C in equation (III)

$\dfrac{2}{\sqrt{v}}=\dfrac{1}{12}x+\dfrac{1}{2}$

We need to calculate the velocity of the particle when x = 5

Put the value in the equation

$\dfrac{2}{\sqrt{v}}=\dfrac{1}{12}\times5+\dfrac{1}{2}$

$\dfrac{2}{\sqrt{v}}=0.916$

$\sqrt{v}=\dfrac{2}{0.9167}$

$v=(2.181)^2$

$v=4.76\ mm/s$

Hence, The velocity of the particle is 4.76 mm/s.