Question

Артист прыгнул на сетку со скоростью 9 м/с до полной остановки сетка растянулась на 1.5 метр.найти ускорение торможения

Answer 1

S=(v^2-V0)\2a
a=(V^2-V0)\2S
a=(0-9)\2*1,5=-9\3=-3 м\с
где минус как раз и показывает противоположное со скоростью движение - т. е. торможение

Answer 2

Calculation of Deceleration

To find the deceleration (or acceleration due to braking), we can use the equation of motion:

v^2 = u^2 + 2as

Where: - v is the final velocity (0 m/s, as the artist comes to a complete stop) - u is the initial velocity (9 m/s, as given in the question) - a is the deceleration (what we need to find) - s is the displacement (1.5 meters, as the net stretches by this amount)

Rearranging the equation, we get:

a = (v^2 - u^2) / (2s)

Substituting the given values, we have:

a = (0^2 - 9^2) / (2 * 1.5)

Simplifying the equation, we get:

a = (-81) / 3

Therefore, the deceleration is -27 m/s^2.

Please note that the negative sign indicates deceleration, as the artist is slowing down.