Two Moving Objects A Distance Apart Montwood High School R. Casao AP Physics 1

Acceleration problems involving two moving objects, either moving in the same direction or in opposite directions, will involve the use of multiple equations to ultimately solve the problem. The time of motion for both objects will be the same. You will need to use a distance equation for each object that includes time. Which distance equation you will use will depend upon the information provided for each object. There is a distance between the two objects at t = 0 s which will be used with the distance equation for each object to describe the simultaneous motion of both objects. If the objects are moving toward each other, the velocity of one object will be positive and the velocity of the second object will be negative. If the objects are moving in the same direction, the velocity of both objects will be positive.

Example: A passenger train, vpt = 25 m/s, is traveling east when the engineer sees the caboose of a freight train 200 m ahead on the same track. The freight train is headed east and vft = 15 m/s. The engineer hits the brakes and the passenger train decelerates at ap = 0.1 m/s2. The freight train continues at constant speed. Take x = 0 m at the location of the front of the passenger train when the engineer applies the brakes. Will there be a collision? If so, where will the collision take place?

If the trains collide, the passenger train will hit the caboose at distance greater than 200 m. The freight train will travel a distance xft. The passenger train will have to travel the 200 m to get where the caboose was and then travel the distance the caboose has moved; xpt = 200 m + xft The time t during which both trains are moving is the same for each train. For each train, use dx = (vi·t) + (½·a·t2) and substitute the appropriate variables. Passenger train: xpt = (25 m/s·t) + (½·-0.1 m/s2·t2) Freight train: xft = (15 m/s·t) + (½·0 m/s2·t2) The velocity for each train is positive because each train is traveling in the same direction. The freight train has constant speed, so the acceleration is 0 m/s2. Substitute the distance for each train into the xpt = 200 m + xft equation and solve for the time t.

(25 m/s·t) + (½·-0.1 m/s2·t2) = 200 m + (15 m/s·t) + (½·0 m/s2·t2) xpt = 200 m + xft (25 m/s·t) + (½·-0.1 m/s2·t2) = 200 m + (15 m/s·t) + (½·0 m/s2·t2) Simplify and combine like terms: (25 m/s·t) + (-0.05 m/s2·t2) = 200 m + (15 m/s·t) (25 m/s·t) - (15 m/s·t) + (- 0.05 m/s2·t2) = 200 m 10 m/s·t + (-0.05 m/s2·t2) = 200 m Quadratic form: a·x2 + b·x + c = 0 -0.05 m/s2·t2 + 10 m/s·t = 200 m -0.05 m/s2·t2 + 10 m/s·t - 200 m = 0 If you don’t already know how to use a calculator to do the quadratic equation for you, learn quickly! Quadratic variables: a = -0.05; b = 10; c = -200 Answers for time: t = 22.54 s and t = 177.46 s The trains will collide 22.54 s after the engineer of the passenger train sees the freight train and applies the brakes.

The passenger train will travel a distance of: xpt = (25 m/s·t) + (½·-0.1 m/s2·t2) = (25 m/s·22.54 s) + (½·-0.1 m/s2·[22.54 s]2) = 538.1 m from x = 0 m. The trains would meet a second time if they were on parallel tracks and continued their motion after the first meeting. Graphing xpt = (25 m/s·t) + (½·-0.1 m/s2·t2) and xft = (15 m/s·t) + (½·0 m/s2·t2) produces this graph (and you can see both times when the trains meet).

The graph meeting point matches the calculated 538.1 m answer.

Passenger train: xpt = (25 m/s·t) + (½·-0.1 m/s2·t2) If the trains are traveling toward each other, the trains will collide somewhere between 0 m and 200 m. Let the passenger train direction be positive and the freight train direction be negative. The distance equation will be 200 m = xpt + xft, but the direction of the freight train is negative, so change the positive sign in front of xft to a negative sign due to its direction. 200 m = xpt - xft Passenger train: xpt = (25 m/s·t) + (½·-0.1 m/s2·t2) Freight train: xft = (-15 m/s·t) + (½·0 m/s2·t2); the velocity of the freight train is also negative due to its direction. Repeating the process with the new equation: 200 m = (25 m/s·t) + (-0.05 m/s2·t2) – [(-15 m/s·t) + (½·0 m/s2·t2)] Simplify: 200 m = (25 m/s·t) + (-0.05 m/s2·t2) + (15 m/s·t) Combine like terms: 200 m = (25 m/s·t) + (15 m/s·t) + (-0.05 m/s2·t2)

Quadratic form: a·x2 + b·x + c = 0 Combine like terms: 200 m = (40 m/s·t) + (-0.05 m/s2·t2) Quadratic form: a·x2 + b·x + c = 0 (-0.05 m/s2·t2) + (40 m/s·t) – 200 m = 0 Quadratic variables: a = -0.05; b = 40; c = -200 Answers for time: t = 5.032 s and t = 794.97 s The trains will collide 5.032 s after the engineer of the passenger train sees the freight train and applies the brakes. The passenger train will travel a distance of: xpt = (25 m/s·t) + (½·-0.1 m/s2·t2) = (25 m/s·5.032 s) + (½·-0.1 m/s2·[5.032 s]2) = 124.5 m from x = 0 m.

The graph meeting point matches the 124.5 m calculated answer.