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 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Sep 23rd 2010, 10:39 PM #1 Junior Member   Join Date: Sep 2010 Posts: 2 Kinematics A train is moving parallel and adjacent to a highway with a constant speed of 21 m/s. Initially a car is 60 m behind the train, traveling in the same direction as the train at 35 m/s and accelerating at 3 m/s^2. What is the speed of the car just as it passes the train? Answer in units of m/s. I have a basic understanding of the problem. I'm trying to find out what Vf of car is when Vf of the car > Vf of the train. But, I'm confused. Is it saying that the Vi of the train is 35 m/s?? Or is the Vi of the train = 0 (Because we have to assume that a train has to stop at one point before it starts again.) I wish I could put my diagram i drew here, but i can't.
 Sep 24th 2010, 06:09 AM #2 Member   Join Date: Jul 2009 Posts: 72 the train doesnt stop the car passes the train because its faster you need to write the equations of motion of the car and the train
 Sep 24th 2010, 08:31 AM #3 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,344 The velocity of the train is constant at 21 m/s. So it's position as a function of time is x_train = x_0 + vt = 60 m + 21m/s x t The car is accelerating, so it's position is determined by: x_car = v_i x t + 1/2 at^2 = 35m/s x t + 1/2 x 3 m/s^2 x t^2 Set these two equations equal, and solve for t. IYou can then put that value for t into an equation for the velocity of the car: v_car = v_i + at

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