The displacement x of a particle varies with time t as $x=a{{e}^{-\alpha t}}+b{{e}^{\beta t}}$, where a, b, α, and β are positive constants. The velocity of the particle will
A. Decrease with time
B. Be independent of α and β
C. Drop to zero when α=β
D. Increase with time
Answer
249.9k+ views
Hint: The rate of change of displacement is called the velocity of the particle. The rate of change of a quantity with respect to time can be found by differentiating the quantity with time. So, first, differentiate the displacement equation to find the velocity. Then find the monotonicity of the velocity.
Complete step-by-step answer:
The displacement of the particle is given by,
$x=a{{e}^{-\alpha t}}+b{{e}^{\beta t}}$
The velocity of a particle having non-linear displacement is given by,
$v=\dfrac{dx}{dt}$........................(1)
Where,
$v$ is the velocity of the particle
$x$ is the displacement of the particle.
So, we can find the velocity of the particle using equation (1),
$v=\dfrac{d}{dt}(a{{e}^{-\alpha t}}+b{{e}^{\beta t}})$
$\Rightarrow v=-a\alpha {{e}^{-\alpha t}}+b\beta {{e}^{\beta t}}$
Hence, the velocity of the particle is given by,
$-a\alpha {{e}^{-\alpha t}}+b\beta {{e}^{\beta t}}$
Now, we need to find the monotonicity of the expression to check the increasing or decreasing nature of the velocity.
We can find the monotonicity of the velocity by differentiating again.
We, we get the following expression by taking the double derivative of displacement,
$\dfrac{dv}{dt}=a{{\alpha }^{2}}{{e}^{-\alpha t}}+b{{\beta }^{2}}{{e}^{\beta t}}$........................(2)
Here, a, b, α, and β are positive quantities.
Also, $a{{\alpha }^{2}}{{e}^{-\alpha t}}$ and $b{{\beta }^{2}}{{e}^{\beta t}}$ cannot be negative quantities.
Hence the entire expression is positive.
If the derivative of expression is positive, then we can say that the expression is increasing.
Hence, we can say that the velocity is increasing.
So, the correct answer is (D).
Note:
Equation (2) actually gives the acceleration of the particle. Acceleration is the rate of change in velocity. As we can see, the acceleration of the particle is always positive. Hence, the velocity of the particle should only increase with time.
Complete step-by-step answer:
The displacement of the particle is given by,
$x=a{{e}^{-\alpha t}}+b{{e}^{\beta t}}$
The velocity of a particle having non-linear displacement is given by,
$v=\dfrac{dx}{dt}$........................(1)
Where,
$v$ is the velocity of the particle
$x$ is the displacement of the particle.
So, we can find the velocity of the particle using equation (1),
$v=\dfrac{d}{dt}(a{{e}^{-\alpha t}}+b{{e}^{\beta t}})$
$\Rightarrow v=-a\alpha {{e}^{-\alpha t}}+b\beta {{e}^{\beta t}}$
Hence, the velocity of the particle is given by,
$-a\alpha {{e}^{-\alpha t}}+b\beta {{e}^{\beta t}}$
Now, we need to find the monotonicity of the expression to check the increasing or decreasing nature of the velocity.
We can find the monotonicity of the velocity by differentiating again.
We, we get the following expression by taking the double derivative of displacement,
$\dfrac{dv}{dt}=a{{\alpha }^{2}}{{e}^{-\alpha t}}+b{{\beta }^{2}}{{e}^{\beta t}}$........................(2)
Here, a, b, α, and β are positive quantities.
Also, $a{{\alpha }^{2}}{{e}^{-\alpha t}}$ and $b{{\beta }^{2}}{{e}^{\beta t}}$ cannot be negative quantities.
Hence the entire expression is positive.
If the derivative of expression is positive, then we can say that the expression is increasing.
Hence, we can say that the velocity is increasing.
So, the correct answer is (D).
Note:
Equation (2) actually gives the acceleration of the particle. Acceleration is the rate of change in velocity. As we can see, the acceleration of the particle is always positive. Hence, the velocity of the particle should only increase with time.
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