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

The subject is Physics, addressing a high school level kinematics problem involving a particle's position under constant acceleration. The example shows how to calculate the particle's position after 5 seconds.

Step-by-step explanation:

The question involves the motion of a particle under a given variable acceleration. Since kinematic equations are required to solve the particle's velocity and position, it is evident that the subject is Physics, specifically mechanics.

Sample Physics Problem

A particle moves in a straight line with an initial velocity of 0 m/s and a constant acceleration of 30 m/s². If x = 0 at t = 0, to find the particle's position at t = 5 s, you would use the kinematic equation:

x(t) = x_0 + v_0t + ½at²

Substituting the given values:

x(5 s) = 0 + 0⋅(5 s) + ½(30 m/s²)⋅(5 s)²

After calculating, the particle's position at t = 5 s is 375 m.

Answer 2

The velocity of the particle is 4.76 mm/s.

Step-by-step 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=(dx)/(dt)`

`dt=(dx)/(v)`....(I)

The acceleration of the particle along a straight line

`a=(dv)/(dt)`

`dt=(dv)/(a)`....(II)

From equation (I) and (II)

`(dx)/(v)=(dv)/(a)`

`a=v(dv)/(dx)`

Put the value of a

`-kv^(2.5)=v(dv)/(dx)`

`(dv)/(dx)=-kv^(1.5)`

On integrating

`(2)/(√(v))=-kx+C`....(III)

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

`(2)/(√(16))=0+C`

`C=0.5`

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

`(2)/(√(4))=-k*6+0.5`

`-k*6=1-(1)/(2)`

`k=-(1)/(12)`

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

`(2)/(√(v))=(1)/(12)x+(1)/(2)`

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

Put the value in the equation

`(2)/(√(v))=(1)/(12)*5+(1)/(2)`

`(2)/(√(v))=0.916`

`√(v)=(2)/(0.9167)`

`v=(2.181)^2`

`v=4.76\ mm/s`

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