Question

In some appropriate units, time (t) and position (x) relation of a moving particle is given by \(t = \alpha x^2 + \beta x\). The acceleration of the particle is :

• A. \( -2\alpha v^3 \)
• B. \( 2\beta v^3 \)
• C. \( -2\beta v^3 \)
• D. \( -2\alpha \frac{v^3}{(2\alpha x + \beta)^2} \)

Hint:

When position is not directly given as a function of time, use the chain rule for differentiation to find velocity and acceleration. Remember that \(v = dx/dt\) and \(a = dv/dt = (dv/dx)(dx/dt) = v (dv/dx)\). Alternatively, \(a = d^2x/dt^2\).

Answer 1

The Correct Option is A

To determine the acceleration of the particle given the relation between time (t) and position (x) as t = αx² + βx, we need to perform the following steps:

1. Differentiate with respect to x:
   The given equation is t = αx² + βx.
   Differentiate both sides with respect to x to find the expression for dt/dx:
   dt/dx = 2αx + β

2. Calculate velocity:
   Velocity (v) is defined as dx/dt. Using the result from step 1, we can write:
   v = 1/(dt/dx) = 1/(2αx + β)

3. Differentiate velocity to find acceleration:
   Acceleration (a) is the derivative of velocity with respect to time, a = dv/dt. Using the chain rule:
   dv/dt = (dv/dx) * (dx/dt)
   Now, first calculate dv/dx:
   v = (2αx + β)^-1
   Differentiate v with respect to x:
   dv/dx = -1 * (2αx + β)^-2 * 2α
   Plug this back into the chain rule expression:
   dv/dt = (-2α/(2αx + β)²) * (dx/dt)

4. Simplify using v:
   Recall that v = 1/(2αx + β), so dx/dt = v.
   Substitute dx/dt in the expression:
   dv/dt = -2αv/(2αx + β)²
   Since v = 1/(2αx + β),
   (2αx + β) = 1/v
   Thus, (2αx + β)² = 1/v²
   Now,
   dv/dt = -2αv * v² = -2αv³

Therefore, the acceleration of the particle is -2αv³.