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Excuse me, math questions:
Let the speed of A be 3x, then the speed of B is 2x.

After acceleration, the speed of A is (3+3* 1/3)x=4x.

The speed of B is (2+2 * 0.25) x = 2.5x..

So after the first meeting, the speed ratio between them is (4): (2.5) = 8: 5.

∵ The first meeting A walked 3/5 of the whole journey, that is, 0.6 B walked 2/5 of the whole journey, that is, 0.4.

Let the distance between AB and AB be s.

Later, A walked 2/5 of the whole journey, which is 0.4.

Similarly, B walked 3/5 of the whole journey, which is 0.6.

The equation 0.4S/4x=(0.6S- 14)/2.5x can be listed.

X on both sides disappears by 0.4S/4=(0.6S- 14)/2.5.

The solution is S=40 (km)

The distance between the two places is 40 kilometers.

Note: (1+ 1/3)*2/3 should be the speed ratio 1/3 between B and A after speeding up. Before increasing the speed, the speed of B is 2/3 of that of A. ..