Given info: The given equation is K=(11−u2/c2−1)mc2 .

The equation for the kinetic energy is given as,

K=(11−u2/c2−1)mc2

Here,

k is the kinetic energy.

m is the mass.

u is the speed.

c is the speed of light.

Rearrange the above equation for u .

Kmc2=(11−u2/c2−1)

Let us assume H=Kmc2 .

H=(11−u2/c2−1)11−u2/c2=1+H1−u2/c2=(11+H)2u2=c2(1−(11+H)2)

Further solve the equation.

u2=c2(1−(11+H)2)u2=c2(H2+2H+1−1(H2+2H+1))u=cH2+2H(H2+2H+1) (1)

Replace H by Kmc2 in above equation.

u=c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1) (2)

Conclusion:

Therefore, the equation for u from the given equation is c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1) .

Given info: The given equation is K=(11−u2/c2−1)mc2 .

From equation (2), the expression for the speed is given as,

u=c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)

From the above expression all the term is positive as well as the expression contains only positive sign so the minimum possible value that the speed can have according to the above expression is zero.

At zero speed the corresponding value of kinetic energy is also zero.

Conclusion:

Therefore, the minimum possible value of speed can be zero and corresponding kinetic energy will also be zero.

Given info: The given equation is K=(11−u2/c2−1)mc2 .

From equation (2), the expression for the speed is given as,

u=c(Kmc2)2+2(Kmc2)((Kmc2)2+2(Kmc2)+1)

The maximum value of speed is equal to the speed of light according to relativistic concept if the speed becomes more than the speed of light then its energy become unstable that would not exist practically.

At this speed of light, the kinetic energy increases without any limit.

Conclusion:

Therefore, the maximum possible value of speed can be speed of light and corresponding kinetic energy will increases without any limit.

Given info: The given equation is K=(11−u2/c2−1)mc2 .

From equation (1), the expression for the speed is given as,

u=c(H)2+2(H)((H)2+2(H)+1)

Write the expression for the acceleration of a particle.

a=d2udt2

Substitute c(H)2+2(H)((H)2+2(H)+1) for u to find a .

a=d2(c(H)2+2(H)((H)2+2(H)+1))dt2=c((H)2+2(H)((H)2+2(H)+1))−1/2(H+1(H+1)4)d(H)dt=cH1/2(H+2)1/2(H+1)2d(H)dt

Replace H by Kmc2 and P by dKdt .

a=cH1/2(H+2)1/2(H+1)2d(Kmc2)dt=1mcH1/2(H+2)1/2(H+1)2dKdt=PmcH1/2(H+2)1/2(H+1)2 (3)

Substitute Kmc2 for H in above equation.

a=Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2 (4)

Conclusion:

Therefore, the equation for the acceleration of the particle as a function of kinetic energy and power input is Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2 .

Given info: The non-relativistic expression for the acceleration is a=P/(2mK)1/2 .

From equation (4), the expression for the acceleration is given as,

a=Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2

At low energy the value of Kmc2 is very small that Kmc2<<<1 . So equation becomes,

a=Pmc(Kmc2)1/2(2)1/2(1)2=P(2mK)1/2

Thus, the limiting form of the expression of acceleration at low energy is P(2mK)1/2 and it is same as that of expression of acceleration of non-relativistic case.

Conclusion:

Therefore, the limiting form of the expression of acceleration at low energy is P(2mK)1/2 and it is same as that of expression of acceleration of non-relativistic case.

Given info: The non-relativistic expression for the acceleration is a=P/(2mK)1/2 .

From equation (4), the expression for the acceleration is given as,

a=Pmc(Kmc2)1/2(Kmc2+2)1/2(Kmc2+1)2

At high energy the value of Kmc2 is very small that Kmc2>>>1 . So equation becomes,

a=Pmc(Kmc2)1/2(Kmc2)1/2(Kmc2)2=Pmc(Kmc2)3=Pm2c5(K)3

Thus, the limiting form of the expression of acceleration at low energy is P(2mK)1/2 and it is same as that of expression of acceleration of non-relativistic case.

Conclusion:

Therefore, the limiting form of the expression of acceleration at high energy is Pm2c5(K)3 .

Given info: The non-relativistic expression for the acceleration is a=P/(2mK)1/2 .

From the answer of part (f) the expression for the acceleration is,

a=Pm2c5(K)3

Here,

p is the power.

m is the mass of a particle.

K is the kinetic energy of the particle.

In part (c), the speed at high energy approaches to the speed of light. But from the acceleration equation if the energy is imparted to the particle at constant input power the acceleration is steeply decreases because the acceleration is inversely proportional to the cube root of the kinetic energy. So at high energy acceleration is very less and the velocity of the particle approaches to a constant value as indicate in part (c).

Conclusion:

Therefore, the acceleration of the particle is very less at high energy that gives the velocity of the particle a constant value.