A red train traveling at 72 km/h and a green train traveling at 144 km/h are hea

Question

A red train traveling at 72 km/h and a green train traveling at 144 km/h are headed toward each other along a straight, level track. When they are 750 m apart, each engineer sees the other's train and applies the brakes. The brakes slow each train at the rate of 1.0 m/s2. Is there a collision? If so, give (a) the speed of the red train, (b) the speed of the green train, and (c) the separation between the trains when they collide (0 m). If not, give (a) the speed of the red train (0 m/s), (b) the speed of the green train (0 m/s), and (c) the separation between the trains when they stop.

Explanation / Answer

Assume red train starts at position 0 and green train starts at position 950(now the graph is in meters). Convert the speeds of the trains to meters per hour Red: 72000m/h Green: 144000m/h Note that the decelleration is in m/s^2, so change the rates to be per second. 60*60 seconds = 1 hour Red: 72000m/3600seconds = 20m/s Green: 144000m/3600seconds = 40m/s So, the function which models the position of red train is x = 0 + 20*t - 1/2 * 1 * t^2 Green train: x = 950 - ( 40t - 1/2 * 1 * t^2 ) Find the intersection between the two trains position and you will have the time for which the trains would collide. Afterwards, we can figure out if the time found is sufficient for the trains to brake. 0 + 20*t - 1/2 * 1 * t^2 = 950 - ( 40t - 1/2 * 1 * t^2 ) 20t - 1/2t^2 = 950 - 40t + 1/2t^2 Set the equation equal to 0. t^2 -60t + 950 = 0 Next, use a method for solving the quadratic in terms of t. If the quadratic formula is used then [-b +- sqrt(b^2 -4ac ) ] / (2a) We have, a = 1, b = -60, c = 950 [-(-60) +- sqrt( (-60)^2 - 4(1)(950) ) ] / [ 2(1) ] [60 +- sqrt( 360 - 4*950) ] / 2 Notice that the radicand(or discriminant in the case of the quadratic) is negative. From this we can be certain that the 0s aren't real numbers. On first inspection, we might conclude that the trains never collide(and rightly so). However, it is always a good idea to check because each train started at a different initial velocity which means one will be braked before the other, and we don't want to add negative distance to that train(in other words, braking can only bring a train/vehicle to 0 velocity, not to a negative velocity). Let us try a different approach(the correct approach). The train which is moving slower is travelling at 20m/s. It decellerates at 1m/s which means it is at 0 when 20 seconds has passed. How far has it travelled? d = vt + 1/2at^2 d = 20*20 -1/2*1*20^2 = 400 - 1/2 *400 = 200, and that was the train that started at position 0. After 20 seconds it is at position 200. What is train green's position after 20 seconds using the distance formula. xNewGreen = 950 - ( 40*20 - 1/2 * 1 * 20^2) = 950 - (800 - 200 ) = 950 - 600 = 350. What speed is the green travelling after the 20 seconds has elapsed? 40 - 20 = 20m/s So, at this time the positions and velocities are as follows. Red: Position = 200, Velocity = 0 Green: Position = 350, Velocity = 20m/s in a negative direction. Then we can repeat again with the green train to find out. When the green train's velocity is 0 will the train be beyond the position of 200 in the negative direction? The green train's velocity will be 0 in another 20 seconds because of the given deceleration. xNewNewGreen = 350 - ( 20*20 - 1/2 * 1 * 20^2 ) = 350 - ( 400 - 200) = 350 - 200 = 150 Therefore, the two trains do collide because the red train is travelling west to east from 0 to 200 and the green train is travelling from east to west from 950 to 150. There is an intersection between 150 and 200 of the two trains' paths.

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