Question

A train slow down from 200km/h With a uniform retardation of 10m/s². How long will it takes to reach 180km/h and what is the total distance covereds?

Answer 1

$$a=-10m/s^2$$

$$a=v'=>v=v_0+at$$

$$v=v_0+at=s'=>s=s_0+v_0t+\dfrac{at^2}{2}$$

Given $v_0=200km/h=\dfrac{200}{3.6}m/s$

$v_1=180km/h=\dfrac{180}{3.6}m/s$

$$v_1=v_0+at_1$$

$$t_1=\dfrac{v_1-v_0}{a}=\dfrac{\dfrac{180}{3.6}m/s-\dfrac{200}{3.6}m/s}{-10m/s^2}=$$

$$=\dfrac{5}{9}s\approx0.556 s$$

If $s_0=0$

$$s_1=\dfrac{200}{3.6}m/s(\dfrac{5}{9}s)+\dfrac{-10m/s^2(\dfrac{5}{9}s)^2}{2}=29.32m$$

It takes $\dfrac{5}{9}s\approx0.556 s$ to reach 180km/h. The total distance covered is $29.32m.$