Question

#Answer the question with steps#

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Answer 1

according to me this is the answer
plz correct me if wrong
comment if correct

Answer 2

We are given displacement of particle as a function of time.

$x=ae^{-\alpha t}+be^{\beta t}$
 
Velocity is the rate of change of displacement with respect to time. So we can differentiate displacement w.r.t time to find velocity.


$v = \frac{dx}{dt}\\ \\ \\ \implies\boxed{v=-a\alpha e^{-\alpha t}+b\beta e^{\beta t}}$

We can see that velocity is a function of time. And it is also dependent on $\alpha$ and $\beta$. 


So, Option (2) is incorrect.


Here only, we can also check with option (3).

Let us put $\alpha = \beta = k$
We have:

$v = -a\alpha e^{-\alpha t}+b\beta e^{\beta t} \\ \\ \implies v = -ake^{-kt} + bke^{kt} \\ \\ \implies v = k(be^{kt}-ae^{-kt})$

Thus, velocity is not zero. 

$\displaystyle \boxed{\begin{minipage}{15 em}\text{EXTRA INFO} \\ \\ \text{If we want velocity to be zero,} \\ \text{ then the additional condition is:} \\ \\ \\ v = 0 \\ \\ \implies k(be^{kt}-ae^{-kt})=0 \\ \\ \implies be^{kt}=ae^{-kt} \\ \\ \implies be^{kt} = \frac{a}{e^{kt}} \\ \\ \implies (e^{kt})^2 = \frac{a}{b} \\ \\ \implies e^{2kt}=\frac{a}{b} \\ \\ \text{Which might not always be true}\end{minipage}}$

Thus, Option (3) is also Incorrect.


Now we are left with Options (1) and (4). Both say opposite things, so we just need to find which one of them is correct. 

Now, from here, I can think of three approaches to solve the question. The first two do the same thing, but the way of thinking is different.


Mathematical Approach


Basically, the concept is that of how we can check if a function is increasing or decreasing. 

Given a function, we find the first derivative. [Remember that the first derivative gives the slope of the function].


If the first derivative [i.e. the slope] is positive, then the function is always increasing. 

If the first derivative is negative, then the function is always decreasing.


To Summarize:

$\mathbb{CONCEPT}\\ \boxed{\begin{minipage}{23em} \text{Consider a function y = f(x)}\\ \\ \\ \text{If }\frac{dy}{dx}=f'(x)>0\implies f(x)\text{ is strictly increasing}\\ \\ \text{If }\frac{dy}{dx}=f'(x)<0\implies f(x)\text{ is strictly decreasing} \end{minipage}}$


Here, we have velocity as a function of time. We will differentiate velocity with respect to time. If the resulting function is positive, then the velocity is always increasing. If the resulting function is negative, the velocity is always decreasing.


$v = -a\alpha e^{-\alpha t}+b\beta e^{\beta t} \\ \\ \implies \frac{dv}{dt} = a\alpha^2e^{-\alpha t}+b\beta^2e^{\beta t}$

Now, exponential functions are always positive. 

For example, if you take a positive number n, and you consider a power $n^{something}$, it will be always positive. 

Again for example, you would never see $2^{something}<0$

Exponential functions with positive base always give positive values.


So we have:

$e^{anything}$ is always positive.

So, $e^{-\alpha t} > 0$ and $e^{\beta t}>0$

And also, we are given that $a, \, \, b, \, \, \alpha, \, \, \text{and} \, \, \beta$ are positive constants.


So, we have:


$\frac{dv}{dt} = a\alpha^2e^{-\alpha t}+b\beta^2e^{\beta t} \ \textgreater \ 0 \\ \\ \\ \implies \frac{dv}{dt} \ \textgreater \ 0 \\ \\ \\ \implies \text{v is increasing}$

Thus, Velocity is always increasing.


Physical Approach



We know that acceleration is the rate of change of velocity with respect to time. So we can find acceleration by differentiating velocity with respect to time. 

The symbol $a$ is already used as a constant. So we will denote acceleration here by $f$


$\displaystyle v = -a\alpha e^{-\alpha t}+b\beta e^{\beta t} \\ \\ \\ \implies f = \frac{dv}{dt} = a\alpha^2e^{-\alpha t}+b\beta^2e^{\beta t}$
As we saw in the Mathematical Approach, the function $\frac{dv}{dt}$ is always positive.


$\frac{dv}{dt} = a\alpha^2e^{-\alpha t}+b\beta^2e^{\beta t} \ \textgreater \ 0 \\ \\ \implies f \ \textgreater \ 0$


Now, since the acceleration is always positive, it means that the change in velocity is always positive. 
We write:

$f = \frac{v-u}{t}$
Where u and v are initial and final velocities respectively.
Since f is always positive, we see that

$\frac{v-u}{t} > 0 \\ \\ \text{Also, time is always positive} \\ \\ \implies v-u > 0 \\ \\ \implies v > u$

That is, in any given interval of time, the final velocity is always greater than the initial velocity.

In other words, Velocity is always increasing with time.


[Option (1) is automatically proved wrong]



There is also a third approach, 

Logical Approach

We have:

$v = -a\alpha e^{-\alpha t}+b\beta e^{\beta t} \\ \\ \\ \implies v = b\beta e^{\beta t} - \frac{a\alpha}{e^{\alpha t}}$

As the value of t increases, $e^{\beta t}$ goes on increasing, and $e^{-\alpha t}$ goes on decreasing. 
So, overall, we are subtracting a decreasing quantity from an increasing quantity. So, velocity must always be increasing.


Thus, Answer is Option (4).