Question

7.
The position x of a particle moving in a straight line,
varies with time tas x = 1/t+3
. The acceleration
of particle is proportional to


​

Answer 1

Explanation:

x = 1/t+3

x = (t+3)^(-1)

v= dx/dt = (-1)(t+3)^(-2)

a = (-1)*(-2)*(t+3)^(-3)

So, a = 2/(t+3)^3

which means that a is proportional to inverse of (t+3)^3 or t^3

Answer 2

Answer:

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

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

Hope this helps.