1.The motion of a body is given by dvdt=6.0−3v\dfrac{dv}{dt} = 6.0 - 3vdtdv=6.0−3v, where vvv is speed in m/s\text{m/s}m/s and ttt in seconds. If the body was at rest at t=0t = 0t=0,Note: More than one option may be correct.a.The terminal speed is 2.0 m/s2.0\,\text{m/s}2.0m/sb.The magnitude of the initial acceleration is 6.0 m/s26.0\,\text{m/s}^26.0m/s2c.The speed varies with time as v=2(1−e−3t) m/sv = 2(1 - e^{-3t})\,\text{m/s}v=2(1−e−3t)m/sd.The speed is 1.0 m/s1.0\,\text{m/s}1.0m/s when the acceleration is half of the initial valueLogin to continueOnly logged in users canattempt or see the solution.