Two points move uniformly along two mutually perpendicular straight lines.

Two points move uniformly along two mutually perpendicular straight lines. Now they are both at the intersection of the straight lines, and in 10 seconds the distance between them will be 1m. Find the speed of each point if one of them travels in 3 s as much as the other travels in 4 s.

Let’s find the ratio of the speeds of the first and second points. The same distance that points travel can be described using the following expressions:

S = 3V1;

S = 4V2;

Let’s equate these expressions and express V1 through V2:

3V1 = 4V2;

V1 = 4V2 / 3

The distances that the points pass in 10 s will be:

S1 = 4V2 / 3 * 10;

S2 = V2 * 10.

Since the straight lines are perpendicular, the distances traveled by the points are the legs of a right-angled triangle with a hypotenuse of 1m. Let’s write this equation:

√ (S1 ^ 2 + S2 ^ 2) = 1;

√ (16V2 ^ 2/9 * 100 + V2 ^ 2 * 100) = 1;

√ (1600V2 ^ 2/9 + V2 ^ 2 * 100) = 1;

√ (1600V2 ^ 2/9 + V2 ^ 2 * 900) = 1;

√ (2500V2 ^ 2/9) = 1;

50V2 / 3 = 1;

V2 = 3/50 m / s

From here we find V1:

V1 = 4V2 / 3 = 4 * 3 / (50 * 3) = 4/50 m / s

Answer: the speed of the first point is 4/50 m / s, the speed of the second is 3/50 m / s.