Question

If the displacement, velocity, and acceleration of a particle at time t be x, v, and f respectively, then which one is true

• A. $f=v^3\frac{d^{2}t}{d{x}^2}$
• B. $f=-v^3\frac{d^{2}t}{d{x}^2}$
• C. $f=v^2\frac{d^{2}t}{d{x}^2}$
• D. $f=-v^2\frac{d^{2}t}{d{x}^2}$

Answer 1

The correct option is B

$f=-v^3\frac{d^{2}t}{d{x}^2}$

Step 1: Given data

Displacement = $x$

Velocity = $v$

Acceleration = $f$

Step 2: Formula used

Velocity can usually be written as:

$v=\frac{dx}{dt}$

Where $x$ is displacement.

Step 3: Calculation for acceleration

Velocity,

$v=\frac{dx}{dt}$

For that reason, $\frac{1}{v}=\frac{dt}{dx}$

Differentiating with respect to 'x"

$-\frac{1}{v^2}\frac{dv}{dx}=\frac{d}{dx}\left(\frac{dt}{dx}\right)$

Multiplying and dividing on LHS, we get

$-\frac{1}{v^2}\left(\frac{dv}{dt}\right)\left(\frac{dt}{dx}\right)=\frac{d^{2}t}{d{x}^2}$

Now, when we express $f$ in terms of velocity,

Since, $\frac{dv}{dt}=Acceleration\left(f\right)$

$-\frac{1}{v^2}\left(f\right)\left(\frac{1}{v}\right)=\left(\frac{d^{2}t}{d{x}^2}\right)$

$f=-v^3\frac{d^{2}t}{d{x}^2}$

Hence, the correct answer is option B.