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Chapter 2: Problem 206

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

Short Answer

Expert verified
The closest answer is (D) \(-2 a v^{3}\), but with a factor \(-\frac{1}{4}\) instead of \(-2\).

Step by step solution

01

Write down the given equation

We have the given relation as \(t = a x^{2} + b x\). In this equation, \(t\) is time, \(x\) is distance, and \(a\) and \(b\) are constants.
02

Derive the velocity

Since velocity \(v\) is the rate of change of distance with respect to time, i.e., the derivative of \(x\) with respect to \(t\), we have to invert the original equation and derive it with respect to \(t\). We can express \(x\) as a function of \(t\) and derive it: \(\frac{dx}{dt} = v = \frac{1}{2\sqrt{a t - b}}\).
03

Derive the acceleration

Acceleration \(a\) is the rate of change of velocity with respect to time, i.e., the second derivative of \(x\) with respect to \(t\). So we have to derive the equation for \(v\) that we found in step 2: \(a = \frac{dv}{dt} = -\frac{\sqrt{a}}{4(t-b)^{3/2}}\). equivalently \(a = -\frac{1}{4}a v^3\).

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