Question

The relation between time $t$ and distance $x$ is $t=a{{x}^{2}}+bx$ where $a$ and $b$ are constants. The acceleration is:
$A)-2ab{{v}^{2}}$
$B)2b{{v}^{2}}$
$C)-2a{{v}^{3}}$
$D)2a{{v}^{2}}$

Answer 1

Acceleration is defined as the change in velocity with respect to time. Velocity is defined as the change in displacement with respect to time. The given relation is differentiated twice with respect to time, to determine the acceleration.
Formula used:
$1)A=\dfrac{dv}{dt}$
$2)v=\dfrac{dx}{dt}$

Complete answer:
We know that acceleration of a body is defined as the change in velocity of the body with respect to time. If $A$ denotes the acceleration of body, then, $A$ is given by
$A=\dfrac{dv}{dt}$
where
$A$ is the acceleration of a body
$v$ is the velocity of the body
$\dfrac{dv}{dt}$ is the change in velocity of the body with respect to time
Let this be equation 1.
Now, we also know that velocity of a body is defined as the change in displacement of the body with respect to time. If $v$ denotes the velocity of a body, then, $v$ is given by
$v=\dfrac{dx}{dt}$
where
$v$ is the velocity of a body
$x$ is the displacement of the body
$\dfrac{dx}{dt}$ is the change in displacement of the body with respect to time
Let this be equation 2.
Coming to our question, we are given that time $t$ and distance $x$ are related by
$t=a{{x}^{2}}+bx$
where $a$ and $b$ are constants
Let this be equation 3.
We are required to determine the acceleration.
Differentiation equation 3 with respect to $t$, we have
$\dfrac{dt}{dt}=\dfrac{d(a{{x}^{2}}+bx)}{dt}\Rightarrow 1=2ax\dfrac{dx}{dt}+b\dfrac{dx}{dt}$
Let this be equation 4.
Using equation 2 and equation 4, we have
$1=2axv+bv\Rightarrow \dfrac{1}{v}=2ax+b$
Let this be equation 5.
Differentiating equation 5 further with respect to $t$, we have
$\dfrac{d}{dt}\left( \dfrac{1}{v} \right)=\dfrac{d(2ax+b)}{dt}\Rightarrow \dfrac{-1}{{{v}^{2}}}\dfrac{dv}{dt}=2a\dfrac{dx}{dt}$
Let this be equation 6.
Using equation 1 and equation 2 in equation 6, we have
$\dfrac{-1}{{{v}^{2}}}\dfrac{dv}{dt}=2a\dfrac{dx}{dt}\Rightarrow \dfrac{-1}{{{v}^{2}}}A=2av\Rightarrow A=-2a{{v}^{3}}$
Therefore, acceleration is equal to $-2a{{v}^{3}}$.

Hence, the correct answer is option $C$.

Note:
Students can also use other methods to solve the given problem. One such method is by combining equation 1 and equation 2, as follows
$A=\dfrac{dv}{dt}=\dfrac{d}{dt}\left( \dfrac{dx}{dt} \right)=\dfrac{{{d}^{2}}x}{d{{t}^{2}}}$
where
$A$ is the acceleration of a body
$v$ is the velocity of the body
$x$ is the displacement of the body
$\dfrac{{{d}^{2}}x}{d{{t}^{2}}}$ is the second derivative of displacement of the body with respect to time
But it needs to be noted that determining acceleration through this method can be tedious.