\nFor the first train: Distance X = velocity * time = y * y = y\u00b2.
\nFor the second train: Distance X = velocity * time = (ky) * (y + 3).
\nSince the distance X is the same:
\ny\u00b2 = ky(y + 3)
\nGiven y is a velocity and time, y > 0. We can divide both sides by y:
\ny = k(y + 3)
\ny = ky + 3k
\ny – ky = 3k
\ny(1 – k) = 3k
\ny = 3k \/ (1 – k)<\/p>\nWe are given that y < 13. So,\n3k \/ (1 - k) < 13\n\nFor the velocity ky and time (y+3) to be positive, k must be positive (since y>0 and y+3>0). Also, for the denominator (1-k) to result in a positive value for y (which must be positive as it’s velocity\/time), (1-k) must be positive, meaning k < 1. So, 0 < k < 1.\n\nSince 0 < k < 1, (1 - k) is positive. We can multiply the inequality by (1 - k) without changing the direction:\n3k < 13(1 - k)\n3k < 13 - 13k\n3k + 13k < 13\n16k < 13\nk < 13\/16.\n\nThus, if y < 13, then k is less than 13\/16.\n<\/section>\n