The distance between the piers A and B is 45 km, from A to B along the course of the rivers the raft departed

The distance between the piers A and B is 45 km, from A to B along the course of the rivers the raft departed, and an hour later a motor boat set off after it, which, having arrived at point B, immediately turned back and returned to A. By that time the raft passed 28 km. Find the speed of the boat in still water if the speed of the river is 4 km / h.

1) How long was the raft on the way? To find the time, the distance traveled must be divided by the speed.

28: 4 = 7 (h)

2) How long was the motor boat on the way? She left an hour later than the raft, which means she was an hour less on the way.

7 – 1 = 6 (h)

3) What is the speed of a motor boat in still water (or the boat’s own speed)?

Let the Own speed of the boat be x km / h, then the speed of the boat along the river is (x + 4) km / h, and the speed of the boat against the river is (x – 4) km / h. The boat covered 45 km both upstream and upstream. It took 45 / (x + 4) hours to move along the river, and 45, (x – 4) hours to move against the current. The boat spent (45 / (x + 4) + 45 / (x – 4)) hours or 6 hours for the entire journey. Let’s make an equation and solve it.

45 / (x + 4) + 45 / (x – 4) = 6;

45 / (x + 4) + 45 / (x – 4) = 6/1 – bring to a common denominator (x – 4) (x + 4) = x ^ 2 – 16; an additional factor for the first fraction (x – 4), for the second fraction – (x + 4), for the third – (x ^ 2 – 16); then we solve without a denominator, since for equal fractions with the same denominator, the numerators are equal;

45 (x – 4) + 45 (x + 4) = 6 (x ^ 2 – 16);

45x – 180 + 45x + 180 = 6x ^ 2 – 96;

6x ^ 2 – 96 = 45x + 45x;

6x ^ 2 – 90x – 96 = 0 – divide term by 6;

x ^ 2 – 15x – 16 = 0;

D = b ^ 2 – 4ac;

D = 15 ^ 2 – 4 * 1 * (- 16) = 225 + 64 = 289; √D = 17;

x = (- b ± √D) / (2a);

x1 = (15 + 17) / 2 = 32/2 = 16 (km / h)

x2 = (15 – 17) / 2 = – 2/2 = – 1 – the speed cannot be negative.

Answer. 16 km / h.