The elevator car rises in 4s uniformly accelerated, reaching a speed of 4m / s. at this speed

| September 7, 2021 | education

The elevator car rises in 4s uniformly accelerated, reaching a speed of 4m / s. at this speed, it moves for 8 seconds, and then 3 seconds – evenly slowed down to a stop. determine the path for the entire time of movement.

t1 = 4 s.

V0 = 0 m / s.

V = 4 m / s.

t2 = 8 s.

t3 = 3 s.

S -?

The path of the elevator car S will be the sum: S = S1 + S2 + S3, where S1 is the movement of the car in time t1, S2 is the movement of the car in time t2, S3 is the movement of the car in time t3.

S1 = a1 * t1 ^ 2/2, a1 = (V – V0) / t1 = V / t1.

S1 = V * t1 ^ 2 / t1 * 2 = V * t1 / 2.

S1 = 4 m / s * 4 s / 2 = 8 m.

S2 = V * t2.

S2 = 4 m / s * 8 s = 32 m.

S3 = (V ^ 2 – V0 ^ 2) / 2 * a3 = V ^ 2/2 * a3.

a3 = (V – V0) / t3 = V / t3.

S3 = V ^ 2 * t3 / 2 * V = V * t3 / 2.

S3 = 4 m / s * 3 s / 2 = 6 m.

S = 8 m + 32 m + 6 m = 46 m.

Answer: the elevator car travels a path S = 46 m.



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