The speed of car B approaching car A = (73.15-40) km/hr = 33.15 km/hr and the distance for car B to catch up with car A = when car B starts.
The lead distance of a car = 40 x 1 1/60 km.
Time required for car B to catch up with car A = distance for car B to catch up with car A/speed for car B to approach car A = 40x11/60/33.15hr = 0.2212hr, and the time required for car B to catch up with car A is exactly equal to the time required for car B to travel from S to T.
S
Distance of t = speed of b train, time required for x b train from s to t = 73.15x0.2212km =16.18km.
Set s
The distance between t is yKM.
Then y/40 = y/73.15+1/60 can be obtained by solving the equation: y= 1.47 (accurate to two decimal places) is S.
The distance of t is1.47km2012-07-23 23: 38: 08 Supplement: The speed of 73. 15km in the title is a bit strange.
Solving the equation is troublesome, so it is omitted. 2012-07-23 23: 52: 21Supplement: Sorry, the solution of the equation should be y = 16.5438+08, so S.
The distance between t should be 16. 18km.