Question

The position x of a particle moving in a straight line varies with time t as x=(t+3)^-1. The acceleration of particle is proportional to

Answer 1

Answer:

Option (1) $(\text{velocity})^{3/2}$

Explanation:

Given, the position x varies with time t as

$x=(t+3)^{-1}$

or, $x=\frac{1}{x+3}$

We know that rate of change of displacement is velocity

i.e. $v=\frac{dx}{dt}$

or, $v=\frac{d}{dt}(t+3)^{-1}$

or, $v=-(t+3)^{-2}$

Also, the rate of change of velocity is accelration

i.e. $a=\frac{dv}{dt}$

or, $a=\frac{d}{dt}[-(t+3)^{-2}]$

$\implies a=2(t+3)^{-3}$

$\implies a=2[(t+3)^{-2}]^{3/2}$

$\implies a=2[-v]^{3/2}$

$\implies a=-2v^{3/2}$

Thus, $a \propto v^{3/2}$

or, $\text{acceleration}\propto (\text{velocity})^{3/2}$

Answer 2

Given that,

The position x of a particle moving in a straight line varies with time t as $x=(t+3)^{-1}$

To find,

The acceleration of the particle.

Solution,

Finding velocity v first.

$v=\dfrac{dx}{dt}\\\\v=\dfrac{d}{dt}(t+3)^{-1}\\\\v=-1(t+3)^{-2}\\\\v=-(t+3)^{-2}$ ......(1)

If a is the acceleration, then,

$a=\dfrac{dv}{dt}\\\\a=\dfrac{d(-(t+3)^{-2})}{dt}\\\\a=-(-2) (t+3)^{-2-1}\\\\a=2(t+3)^{-3}\\\\\implies a=[2(t+3)^{-2}]^{3/2}\\\\\text{from equation (1)}\\\\a=2(-v)^{3/2}\\\\\implies a\propto v^{3/2}$

So, the correct option is (a).