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

Answer:

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

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,

$a=\dfrac{v-u}{t}\\\\a=\dfrac{0-18}{15}\\\\a=-1.2\ m/s^2$

Using third equation of motion,

$v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-18^2}{2\times (-1.2)}\\\\d=135\ m$

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