The velocity of a particle moving along a straight line decreases linearly with

Question

The velocity of a particle moving along a straight line decreases linearly with its position coordinate s from 30 ft/sec at s = 0 to a value approaching zero as s approaches 60 ft. Show that the particle never reaches s = 60 ft, and find the acceleration a when s = 40 ft. Answer a = -5 ft/sec^2. The velocity of a particle moving along a straight line decreases linearly with its position coordinate s from 30 ft/sec at s = 0 to a value approaching zero as s approaches 60 ft. Show that the particle never reaches s = 60 ft, and find the acceleration a when s = 40 ft. Answer a = -5 ft/sec^2.

Explanation / Answer

v(s) = -ks + b, k, b = constant use v(0) = 30 ft/s, 30 = b use v(60) = 0, 0 = -60k + b, so k = 0.5 v = 30 - 0.5s ds/dt = 30 - 0.5s ds/(30 - 0.5s) = dt integrate, ln|30 - 0.5s| = -0.5t + lnC 30 - 0.5s = C*e^(-0.5t) s = 60 - 2C*e^(-0.5t) note C*e^(-0.5t) > 0, so s < 60 ft, it means the particle never reaches s = 60 ft. v = 30 - 0.5 s a = dv/dt = -0.5 ds/dt = -0.5 v = -0.5 *(30 - 0.5 s) when s = 40 ft, a = -0.5*(30 - 0.5^40) = -5 ft/s^2

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