(2) An ant crawls in a straight line with a constant speed v = 12 cms. One second after it passes a beetle, the beetle begins crawling in the same direction, starting from rest and moving with a constant acceleration of 4 cm/s2. (a) When does the beetle catch up to the ant? (b) How far has each traveled? Solution here, let the time taken to catch up be t and distance travelled by beetle be x for ant (x ) = v * t x = 12 * t ...(1) and x + 12 = 0 + 0.5 * a * t^2 x + 12 = 0 + 0.5 * 4 * t^2 ....(2) from (1) ans (2) x = 82.5 cm t = 6.87 s the time taken is 6.87 s b) the distance travelled by ant is 82.5 cm the distance travelled by bettle = x + 12 = 94.5 cm .