40 seconds after the departure of the motor ship, a boat with constant acceleration, the modulus of which

| July 1, 2021 | education

40 seconds after the departure of the motor ship, a boat with constant acceleration, the modulus of which was 0.5 m / s2, set off after him. The motor ship moved uniformly with a speed module of 18 km / h. The boat will catch up with the ship, being on the way during the time t1 = 1.

Given: Δt (difference between the beginning of the movement of the motor ship and the boat) = 40 s; ak (constant boat acceleration module) = 0.5 m / s2; Vt (constant speed of the ship) = 18 km / h (in SI Vt = 5 m / s).

The time during which the boat will be able to catch up with the motor ship, we determine from the equality: Vt * (t + Δt) = S = a * t ^ 2/2.

5 * (t + 40) = 0.5 * t ^ 2/2.

5t + 200 = 0.25t ^ 2.

0.25t ^ 2 – 5t – 200 = 0 | * four.

t2 – 20t – 800 = 0.

By Vieta’s theorem: t1 = 40 s (true); t2 = -20 s (incorrect).

Answer: 3) the boat will catch up with the motor ship in 40 s.



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