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The velocity of a particle traveling along a straight line is \(v=v_{0}-k s\), where \(k\) is constant. If \(s=0\) when \(t=0\), determine the position and acceleration of the particle as a function of time.

Short Answer

Expert verified
The position of the particle at time \(t\) is given by \(s = v_{0}/k - e^{(-kt)}\) and the acceleration at time \(t\) is given by \(a(t) =ke^{(-kt)}\).

Step by step solution

01

Solving for Position

Re-arrange the equation to get \(ds = dt( v_{0}-k s)\), then integrate both sides. The integral of the left side will give the position \(s\) and the right side will give the expression for time \(t\), leading to: \(s = v_{0}/k - e^{(-kt)}\).
02

Solving for Acceleration

The acceleration is given by \(a(t) = dv/dt = -k ds/dt\). By substituting the equation for position obtained from step 1 into this equation and differentiating we get: \(a(t) =ke^{(-kt)}\).

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Most popular questions from this chapter

The roller coaster car travels down the helical path at constant speed such that the parametric equations that define its position are \(x=c \sin k t, y=c \cos k t, z=h-b t\), where \(c, h\), and \(b\) are constants. Determine the magnitudes of its velocity and acceleration.

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