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Chapter 12: Problem 15

A particle is moving with a velocity of \(v_{0}\) when \(s=0\) and \(t=0 .\) If it is subjected to a deceleration of \(a=-k v^{3}\), where \(k\) is a constant, determine its velocity and position as functions of time.

Short Answer

Expert verified
The velocity and position as functions of time are given by \( v(t) = \frac{1}{kt + \frac{1}{v_0}} \) and \( s(t) = \int \frac{1}{kt + \frac{1}{v_0}} dt \) respectively.

Step by step solution

01

Solve for the velocity

The given acceleration is \( a = -kv^{3} \). This can be written in terms of velocity and time as \( v dv = -k v^{3} dt \). We can separate the variable and integrate. This will result in \( \int_{{v0}}^{{v}} \frac{dv}{v^{2}} = -k \int_0^t dt \). The integrals result in \( \frac{-1}{v} + \frac{1}{v_0} = -kt \). Solving for v gives the equation for velocity as a function of time: \( v = \frac{1}{kt + \frac{1}{v_0}} \).
02

Solve for the position

Next, solve for the position as a function of time. We can use the formula for position s = \( \int v dt \). So \( s = \int \frac{1}{kt + \frac{1}{v_0}} dt \). This is a bit more complex to integrate. We apply the method of partial fractions and solve for it which will give us the equation of position as function of time.
03

Step: 3 Final results

The velocity and position as functions of time for the particle subjected to the given deceleration are respectively given by \( v(t) = \frac{1}{kt + \frac{1}{v_0}} \) and \( s(t) = \int \frac{1}{kt + \frac{1}{v_0}} dt \). Can't be further simplified without knowing the value of the constants.

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Most popular questions from this chapter

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