Two particles of mass m each are connected by q15 a mass less spring of length / and spring constant k. the system is lying (in the relaxed state) on a frictionless horizontal table, a very short impulse p is applied to the first mass (along the axis of the system) at t = 0. when would the second mass first come to rest? 2m (b) t = 27t. k 2m (a) t = t m (c) t t 2k m (d) t t, k

Answer: it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.explanation:   we have equation of motion ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/sso it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.

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