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Step-by-Step Solution
Step 1: Write down the given displacement
The displacement of the particle is given by
$x = a e^{-\alpha t} + b e^{\beta t}$,
where $a$, $b$, $\alpha$, and $\beta$ are positive constants.
Step 2: Differentiate to find the velocity
Velocity $v$ is the first derivative of displacement with respect to time $t$:
$$
v = \frac{dx}{dt}
= \frac{d}{dt}(a e^{-\alpha t} + b e^{\beta t})
= -\alpha\,a\,e^{-\alpha t} + \beta\,b\,e^{\beta t}.
$$
Step 3: Check how the velocity changes with time
To see whether $v$ goes on increasing with time, we look at its derivative with respect to time:
$$
\frac{dv}{dt}
= \frac{d}{dt}\bigl(-\alpha\,a\,e^{-\alpha t} + \beta\,b\,e^{\beta t}\bigr).
$$
Compute each term separately:
$$
\frac{d}{dt}\bigl(-\alpha\,a\,e^{-\alpha t}\bigr)
= -\alpha\,a \Bigl(\frac{d}{dt} e^{-\alpha t}\Bigr)
= -\alpha\,a \bigl(-\alpha\,e^{-\alpha t}\bigr)
= \alpha^2\,a\,e^{-\alpha t},
$$
$$
\frac{d}{dt}(\beta\,b\,e^{\beta t})
= \beta\,b \Bigl(\frac{d}{dt} e^{\beta t}\Bigr)
= \beta\,b \bigl(\beta\,e^{\beta t}\bigr)
= \beta^2\,b\,e^{\beta t}.
$$
Hence,
$$
\frac{dv}{dt}
= \alpha^2\,a\,e^{-\alpha t} + \beta^2\,b\,e^{\beta t}.
$$
Step 4: Conclude that velocity increases with time
Since $a>0$, $b>0$, $\alpha>0$, $\beta>0$, and exponentials are always positive, we have
$\alpha^2\,a\,e^{-\alpha t} > 0$ and $\beta^2\,b\,e^{\beta t} > 0$ for all $t$. Therefore,
$$
\frac{dv}{dt} = \alpha^2\,a\,e^{-\alpha t} + \beta^2\,b\,e^{\beta t} > 0
$$
at all times. This means $v$ is an increasing function of time.
Answer
The velocity of the particle goes on increasing with time.
