A 2.00 kg box is moving to the right with a speed of 9.00 m/s, on a horizontal frictionless surface. At t = 0, a horizontal force is applied to the box. The force is directed to the left and has a magnitude F(t) = (6.00 N/s^2)t^2A) What distance does the box move from its position at t = 0 before its speed is reduced to zero?B) If the force continues to be applied what is the speed of the box at t = 3.00s.

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Answersystem;

Distance covered = speed x timethe distance traveled going against the stream = distance traveled with the stream.therefore it is the speed and the time that are the variables.as he travels the same distance both ways, speed x time is the same for each way.going upstream his speed is 4 kph; going downstream his speed is 6 kph.therefore we have two simultaneous equations: time upstream x 4 = time downstream x 6time upstream + time downstream = 60 minutestherefore we have the equation: time upstream = 6/10 hour, time downstream therefore = 4/10 hour.distance traveled at 4km/hr in 6/10 hour = 24/10 = 2.4km.check: distance traveled at 6km/hr in 4/10 hour = 24/10 = 2.4km.