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Chapter 6: Problem 80

A locomotive accelerates a freight train of total mass \(M\) from rest, applying constant power \(P\). Determine the speed and position of the train as functions of time, assuming all the power goes to increasing the train's kinetic energy.

Short Answer

Expert verified
The speed of the train as a function of time is \(V = \left(\frac{2PT}{M}\right)^{1/2}\) and the position of the train as a function of time is \(x(t) = \left(\frac{2P}{3M}\right)t^{3/2}\).

Step by step solution

01

Express the power

Firstly, as given, the power \(P\) is constant and all of it converts to kinetic energy. The power can be expressed in the form of rate of change of kinetic energy, which is \(P = \frac{d KE}{dt} = \frac{d}{dt}(\frac{1}{2} M v^2)\)
02

Derive the expression for velocity

By using the chain rule in calculus, we can rewrite the derivative to isolate the velocity: \(P = Mv \frac{dv}{dt}\). We can rearrange it to get an expression for the rate of change of velocity in terms of the power and the mass: \(\frac{dv}{dt} = \frac{P}{Mv}\). Then, we can integrate both sides from time \(t = 0\) to \(t = T\) and speed from \(v = 0\) to \(v = V\). The result will be : \(V = \left(\frac{2PT}{M}\right)^{1/2}\)
03

Derive the expression for position

To find the position as a function of time, we need to integrate the velocity with respect to time. The position \(x(t)\) is the integral of the speed from \(t = 0\) to \(t = T\). The integral of \(V\) with respect to \(t\) would give \(x(T) = \left(\frac{2P}{3M}\right)T^{3/2}\). Thus, the position of the train as a function of time is \(x(t) = \left(\frac{2P}{3M}\right)t^{3/2}\)

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