Question

Скорость поезда за 20 с уменьшилась с 72 до 0 км/ч.Найти тормозной путь и тормозное ускорение

Answer 1

V0=72км/ч=72×1000/3600=20м/с
Ускорение
a=(v-v0)/t
a=(0-20)/20=-1м/с
Тормозной путь
S=v0t-at^2/2
S=20×20-1×400/2=200м

Answer 2

Calculation of Braking Distance and Deceleration

To calculate the braking distance and deceleration, we need to use the equation of motion:

S = V0t + (1/2)at^2

Where: - S is the braking distance - V0 is the initial velocity - t is the time taken to come to a stop - a is the deceleration

Given that the initial velocity (V0) is 72 km/h and the time taken to come to a stop (t) is 20 seconds, we can substitute these values into the equation to find the deceleration (a) and the braking distance (S).

Let's calculate the deceleration first:

0 = 72 * (20) + (1/2) * a * (20)^2

Simplifying the equation:

0 = 1440 + 200a

Rearranging the equation:

200a = -1440

a = -1440 / 200

a = -7.2 m/s^2

The deceleration is -7.2 m/s^2.

Now, let's calculate the braking distance:

S = 72 * (20) + (1/2) * (-7.2) * (20)^2

Simplifying the equation:

S = 1440 - 1440

S = 0

The braking distance is 0 meters.

Therefore, the braking distance is 0 meters and the deceleration is -7.2 m/s^2.