Question

Un tren va a una velocidad de 18m/s frena y se detiene en 15s calcula su aceleracion y la distancia recorrida al frenar

Answer 1

Acceleration is -1.2 m/s² and distance covered is 135 m.

Step-by-step explanation:

A train going at a speed of 18m / s brakes and stops in 15s calculates its acceleration and the distance traveled when braking

Given that,

The initial speed of the train, u = 18 m/s

Final speed, v = 0

Time, t = 15 s

We need to find acceleration and distance traveled when braking. Let a is acceleration and distance traveled.

Acceleration,

`a=(v-u)/(t)\\\\a=(0-18)/(15)\\\\a=-1.2\ m/s^2`

Using third equation of motion,

`v^2-u^2=2ad\\\\d=(v^2-u^2)/(2a)\\\\d=(0^2-18^2)/(2* (-1.2))\\\\d=135\ m`

Hence, acceleration is -1.2 m/s² and distance covered is 135 m.