Question

The relation between time \( t \) and distance \( x \) is \( t = \alpha x^2 + \beta x \), where \( \alpha \) and \( \beta \) are constants. The relation between acceleration \( a \) and velocity \( v \) is:

• A. \( a = -2 \alpha v^3 \)
• B. \( a = -5 \alpha v^5 \)
• C. \( a = -3 \alpha v^2 \)
• D. \( a = -4 \alpha v^4 \)

Answer 1

The Correct Option is A

To establish the relationship between acceleration \( a \) and velocity \( v \), we begin with the provided equation relating time \( t \) and distance \( x \):

\(t = \alpha x^2 + \beta x\)

The objective is to derive expressions for velocity and acceleration. Velocity \( v \) is defined as:

\(v = \frac{dx}{dt}\)

To compute the derivative, first express \( x \) as a function of \( t \):

\(\frac{dt}{dx} = 2\alpha x + \beta\)

Consequently, velocity \( v \) can be expressed as:

\(v = \frac{dx}{dt} = \frac{1}{\frac{dt}{dx}} = \frac{1}{2\alpha x + \beta}\)

Next, we determine acceleration \( a \), which is the rate of change of velocity with respect to time:

\(a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \cdot \frac{dv}{dx}\)

Applying the chain rule, we differentiate \( v \) with respect to \( x \):

\(v = \frac{1}{2\alpha x + \beta} \implies \frac{dv}{dx} = -\frac{2\alpha}{(2\alpha x + \beta)^2}\)

Substituting the expressions for \( \frac{dv}{dx} \) and \( v \) yields \( a \):

\(a = v \cdot \frac{dv}{dx} = \left(\frac{1}{2\alpha x + \beta}\right) \cdot \left(-\frac{2\alpha}{(2\alpha x + \beta)^2}\right) = -\frac{2\alpha}{(2\alpha x + \beta)^3}\)

Substitute \( v = \frac{1}{2\alpha x + \beta} \) into the acceleration equation:

\(a = -2\alpha \left(\frac{1}{2\alpha x + \beta}\right)^3 = -2\alpha v^3\)

The established relationship between acceleration \( a \) and velocity \( v \) is:

\(a = -2 \alpha v^3\)

The definitive result is \(a = -2 \alpha v^3\).