A car leaves the city at a speed of 18m / s, after 20 minutes a second car leaves after it, at what speed did the second car move if it caught up with the first one an hour after the start of its movement?
V1 = 18 m / s.
t1 = 20 min = 1200 s.
t2 = 1 h = 3600 s.
With uniform rectilinear movement, the path S traversed by the body is determined by the formula: S = V * t, where V is the speed of movement, t is the time of movement.
Before the meeting, the first car covered the path S1 = V1 * t, where V1 is the speed of the first car, t is the travel time of the first car before the meeting.
t = t1 + t2.
S1 = V1 * (t1 + t2).
Before the meeting, the second car covered the path S2 = V2 * t2, where V2 is the speed of the second car, t2 is the travel time of the second car before the meeting.
S1 = S2.
V1 * (t1 + t2) = V2 * t2.
The speed of the second car V2 will be determined by the formula: V2 = V1 * (t1 + t2) / t2.
V2 = 18 m / s * (1200 s + 3600 s) / 3600 s = 24 m / s.
Answer: the speed of the second car is V2 = 24 m / s.
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