Question

The motion of a body is given by the equation dv(t)/dt=6.0-3v(t),where v(t) is speed in m/s and t in sec.If body was at rest at t=0.

Answer 1

$\frac{dv}{dt} = 6-3v \\ \frac{dv}{6-3v} = dt \\ ln(6-3v)/(-3) = t + c$

At t=0, v=0
Therefore c = ln(6)

On simplifying it gives $v = 2(1-e^{-3t})$

Answer 2

Answer:

Explanation: (B) The terminal speed is the constant speed when acceleration is zero. Thus 0 = 6.0-3v v = 2 m/s (C) At t=0, v=0; therefore initial acceleration a = 6.0-0 = 6.0 m/s2 (D) Acceleration is half that of initial value = 6.0/2 = 3.0m/s2 This gives 3 = 6-3v(t) or v(t) = 1m/s D) using integration v=2(1-e^-3t)

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