Question

Чему равен тормоозной путь автобуса, двигающегося со скоростью 54км/ч, если его ускорение равно 4,5м/с в квадрате?

Answer 1

V0=54км/ч=54×1000/3600=15м/с
Тормозной путь
S=v0^2/2a
S=15^2/2×4,5=25м
Ответ 25м

Answer 2

Calculation of Braking Distance

To calculate the braking distance of a bus moving at a speed of 54 km/h with an acceleration of 4.5 m/s^2, we need to use the equation of motion:

v^2 = u^2 + 2as

Where: - v is the final velocity (0 m/s, as the bus comes to a stop) - u is the initial velocity (54 km/h, which needs to be converted to m/s) - a is the acceleration (-4.5 m/s^2, as it is deceleration in this case) - s is the distance traveled during deceleration (braking distance)

First, let's convert the initial velocity from km/h to m/s:

54 km/h = 54 * (1000/3600) m/s = 15 m/s.

Now, we can substitute the values into the equation:

0^2 = 15^2 + 2 * (-4.5) * s

Simplifying the equation:

0 = 225 - 9s

Rearranging the equation to solve for s:

9s = 225

s = 225 / 9

s = 25 meters

Therefore, the braking distance of the bus, moving at a speed of 54 km/h with an acceleration of 4.5 m/s^2, is 25 meters.