Question

Initial acceleration of a particle moving in a straight line is a and initial Velocity is zero. The acceleration reduces continuously to half in every t seconds. The terminal Velocity of the particle is

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Answer 1

Given that,

Initial acceleration of a particle moving in a straight line is a.

Initial velocity is zero.

We know that,

The rate of change of acceleration is directly proportional to the acceleration.

$\dfrac{da}{dt}\propto a$

$\dfrac{da}{a}=kdt$....(I)

We need to calculate the terminal velocity of the particle

Using equation (I)

$\dfrac{da}{a}=-kdt$

Negative sign shows the decrease of acceleration with time.

Where, k = proportional constant

$\int_{a_{0}}^{a}{\dfrac{da}{a}}=-\int_{0}^{t}{kdt}$

$\ln\dfrac{a}{a_{0}}=-kt$

$a=a_{0}e^{-kt}$.....(II)

The terminal velocity is

$v(t)=\int_{0}^{t}{adt}$

Put the value into the formula

$v(t)=\int_{0}^{t}{a_{0}e^{-kt} dt}$

$v(t)=-\dfrac{a_{0}}{k}({e^{-kt}})_{0}^{t}$

$v(t)=-\dfrac{a_{0}}{k}(e^{-kt}=e^{0})$

$v(t)=-\dfrac{a_{0}}{k}(\dfrac{1}{e^{kt}}-1)$

The terminal velocity at t= ∞

$v(t)=-\dfrac{a_{0}}{k}$......(III)

We use $t=t_{0}$ and $a=\dfrac{a_{0}}{2}$

Put the value in equation (I)

$\dfrac{a_{0}}{2}=a_{0}e^{-kt_{0}}$

$2=e^{-kt_{0}}$

$k=\dfrac{ln 2}{t_{0}}$

Put the value int equation (III)

$v(t)=--\dfrac{a_{0}t_{0}}{\ln2}$

Hence,The terminal velocity of the particle is $-\dfrac{a_{0}t_{0}}{\ln2}$