Buggy 1 has a velocity of 3.1 m/s Buggy 2 has a velocity of 5.3 m/s

Determine how far apart you'd have to set them so that they would collide head on with one another EXACTLY at 7 seconds.

Graph the collision on a position-time graph.

3.1*7+5.3*7=21.7+37.1=58.8m

Explanation:

If buggies don't have an acceleration,

The distance of the buggy 1 can travel within a 7 seconds = 3.1ms^-1 *7s = 21.7m

The distance of the buggy 2 can travel within a 7 seconds = 5.3ms^-1*7s = 37.1m

If they are coming in opposition direction. The distance between them represented by the sum of distances between meeting point & buggies

1/0.14nm = 10^10/1.4 a lot

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this