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 :
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- \( -2\alpha v^3 \)
- \( 2\beta v^3 \)
- \( -2\beta v^3 \)
- \( -2\alpha \frac{v^3}{(2\alpha x + \beta)^2} \)
The Correct Option is A
Solution and Explanation
To determine the particle's acceleration given the relationship \(t = \alpha x^2 + \beta x\) between time (\(t\)) and position (\(x\)), follow these steps:
Differentiate with respect to \(x\):
From the equation \(t = \alpha x^2 + \beta x\), differentiate both sides with respect to \(x\) to obtain \(dt/dx\):
\(dt/dx = 2\alpha x + \beta\)Calculate velocity:
Velocity (\(v\)), defined as \(dx/dt\), can be expressed using the result from step 1 as:
\(v = 1/(dt/dx) = 1/(2\alpha x + \beta)\)Differentiate velocity to find acceleration:
Acceleration (\(a\)) is the time derivative of velocity, \(a = dv/dt\). Apply the chain rule:
\(dv/dt = (dv/dx) \cdot (dx/dt)\)
First, calculate \(dv/dx\) by differentiating \(v = (2\alpha x + \beta)^{-1}\) with respect to \(x\):
\(dv/dx = -1 \cdot (2\alpha x + \beta)^{-2} \cdot 2\alpha = -2\alpha/(2\alpha x + \beta)^2\)
Substitute this back into the chain rule expression:
\(dv/dt = (-2\alpha/(2\alpha x + \beta)^2) \cdot (dx/dt)\)Simplify using \(v\):
Since \(v = 1/(2\alpha x + \beta)\), we have \(dx/dt = v\). Substitute this into the acceleration expression:
\(dv/dt = -2\alpha v/(2\alpha x + \beta)^2\)
From \(v = 1/(2\alpha x + \beta)\), it follows that \((2\alpha x + \beta) = 1/v\), and therefore \((2\alpha x + \beta)^2 = 1/v^2\). Substitute this into the acceleration equation:
\(dv/dt = -2\alpha v \cdot v^2 = -2\alpha v^3\)
Consequently, the particle's acceleration is \(-2\alpha v^3\).