Question

поезд идущий со скоростью 15 м/с,остановился через 20 с после начала торможения. Считая,что торможение происходило с постоянным ускорением,определите перемещение поездаза 20 с

Answer 1

Дано: Vo=15м/с

V=0м/с

t=20с

S-?

S=(V^2-Vo^2)/2a

a=(V-Vo)/t

a=15/20=0,75м/с2

S=15*15/2*0,75=150м

 

Answer 2

Problem Analysis

To determine the displacement of the train over 20 seconds while decelerating at a constant rate, we can use the equations of motion. The initial velocity of the train is 15 m/s, and it comes to a stop after 20 seconds.

Solution

The displacement of the train can be calculated using the equation: s = ut + (1/2)at^2

Where: - s = displacement - u = initial velocity - t = time - a = acceleration

Given: - Initial velocity (u) = 15 m/s - Time (t) = 20 s - Acceleration (a) = ?

First, we need to find the acceleration using the equation: v = u + at

Where: - v = final velocity (0 m/s, as the train comes to a stop)

Rearranging the equation to solve for acceleration: a = (v - u) / t

Substitute the known values: a = (0 - 15) / 20

Calculation

a = -0.75 m/s^2

Now, we can use the equation of motion to find the displacement: s = ut + (1/2)at^2

Substitute the known values: s = (15 * 20) + (1/2)(-0.75)(20^2)

Final Calculation

s = 300 + (-150) s = 150 m

Answer

The displacement of the train over 20 seconds while decelerating at a constant rate is 150 meters.