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
[tex]a=-kv^{2.5}[/tex]
Here, k = constant
v= velocity
Th velocity of the particle along straight line
[tex]v=\dfrac{dx}{dt}[/tex]
[tex]dt=\dfrac{dx}{v}[/tex]....(I)
The acceleration of the particle along a straight line
[tex]a=\dfrac{dv}{dt}[/tex]
[tex]dt=\dfrac{dv}{a}[/tex]....(II)
From equation (I) and (II)
[tex]\dfrac{dx}{v}=\dfrac{dv}{a}[/tex]
[tex]a=v\dfrac{dv}{dx}[/tex]
Put the value of a
[tex]-kv^{2.5}=v\dfrac{dv}{dx}[/tex]
[tex]\dfrac{dv}{dx}=-kv^{1.5}[/tex]
On integrating
[tex]\dfrac{2}{\sqrt{v}}=-kx+C[/tex]....(III)
Put the value of x = 0 and v = 16 mm/s in equation (III)
[tex]\dfrac{2}{\sqrt{16}}=0+C[/tex]
[tex]C=0.5[/tex]
Now put the value of x = 6mm and v = 4 mm in the equation (III)
[tex]\dfrac{2}{\sqrt{4}}=-k\times6+0.5[/tex]
[tex]-k\times6=1-\dfrac{1}{2}[/tex]
[tex]k=-\dfrac{1}{12}[/tex]
Put the value of k and C in equation (III)
[tex]\dfrac{2}{\sqrt{v}}=\dfrac{1}{12}x+\dfrac{1}{2}[/tex]
We need to calculate the velocity of the particle when x = 5
Put the value in the equation
[tex]\dfrac{2}{\sqrt{v}}=\dfrac{1}{12}\times5+\dfrac{1}{2}[/tex]
[tex]\dfrac{2}{\sqrt{v}}=0.916[/tex]
[tex]\sqrt{v}=\dfrac{2}{0.9167}[/tex]
[tex]v=(2.181)^2[/tex]
[tex]v=4.76\ mm/s[/tex]
Hence, The velocity of the particle is 4.76 mm/s.
