Question

In the one dimension motion of a particle, the relation between position x and time t is given by x^2 + 2x = t (here x>0). Choose the correct statement.

a) the retardation of the particle is 1/4(x+1)^3
b) the uniform acceleration of the particle is 1/(x+1)^3
c) the uniform velocity of the particle is 1/(x+1)^3
d) the particle has a variable acceleration of 4t+ 6

Answer 1

Answer:

Option (a) the retardation is 1/4(x+1)^3

Explanation:

The relation between position x and time t is given as

$x^2+2x=t$

We know that rate of change of displacement is equal to velocity

i.e. dx/dt = v

And, rate of change of velocity is acceleration

i.e. dv/dt = a

Also, a = v(dv/dx)      

Differentiating the given relation in x and t, with respect to t

$2x\frac{dx}{dt} +2\frac{dx}{dt} =1$

or, $\frac{dx}{dt} =\frac{1}{2(x+1)}$

or, $v=\frac{1}{2(x+1)}$

Again differentiating the above w.r.t. x

$\frac{dv}{dx} =-\frac{1}{2(x+1)^2}$

⇒ $v\frac{dv}{dx} =-\frac{1}{2(x+1)}\times\frac{1}{2(x+1)^2}$

or, $a=-\frac{1}{4(x+1)^3}$

The negative sign indicates retardation

Therefore, the retardation is 1/4(x+1)³

Answer 2

hope it helps you to get the correct solution