1. The problem statement, all variables and given/known data Problem 5: Collision course A huge tractor and a Tesla full of school children come driving along a winding mountain road, in opposite directions. The tractor has a speed of 40.0km/h and the car zooms along with 80.0 km/h. The Tesla suddenly comes around a corner, sees the tractor, and they both immediately start braking, both with constant accelerations of 5.00 m/s2 (opposite to their directions of motion). a) If the initial distance between the two is 60.0 m, do they hit each other? If so, where, and with what relative speed on impact? If not, what is the distance between the two when they both stop? In fact, it takes both of them 0.50 s to react to seeing each other, so they only start braking 0.50s after the car comes round the corner. b) What is the answer to the questions in a) in this case? As it happens the tractor driver is looking the other way, and doesn’t brake at all, but continues with his original speed. c) What should the acceleration a of the Tesla be, to have time to stop before being hit by the tractor (still including the 0.5 s delay from question b))? 2. Relevant equations 1. x = (a/2)t^2 + vi*t (+xi) 2. x = (v^2-vi^2)/2a 3. v = vi + at (this might be relevant?) 3. The attempt at a solution What I have done so far: I have set the tractor's direction to be positive. Tractor = A Tesla = B vA = 40km/h = 11.1m/s vB = -80km/h = -22.2m/s aA = -5m/s^2 aB = 5m/s^2 x = 60m (distance from each other) Since vA < vB, we know that the tractor will stop before the Tesla will hit it. So vA-finale = 0m/s Then I have made a positional function for both cars. xA = (-vA^2)/2a xB(t) = (a/2)t^2 + vB*t + x Then I set xA = xB(t) (-vA^2)/2a) = (a/2)t^2 + vB*t + x (-11.1)^2/10 = (5/2)t^2 - 22.2t + 60 => (5/2)t^2 - 22.2t + 47.7 = 0 Which gives me t = 3.6 or t = 5.2 (but I assume that the only one valid is t = 3.6?) Since that is the time it takes before they hit each other? Then I insert t = 3.6 into xB(t) to find the position of Tesla (to find out where they "crashed") x = 12.5m Does that mean that he drove (60m - 12.5m) = 47.5m before they crashed? (which means that the tractor drove 12.5m). It´s here I start getting confused. Which equation do I use now to find the speed on impact? V = Vi + at? Where I put in 22.2 m/s for Vi and -5m/s^2 for a, and 3.6 for t? Or do I have to use an equation which includes the position as well? Sorry if my equations are littlebit hard to read (I don´t know the codes to get them look "normal") I hope some of you might help me out! Would love some tips for b and c as well.