Originally Posted by

**Jonroberts74** I solved the problem but I want to make sure I did it correctly.

starting at time t=0, a train is halted by a braking deceleration of $\displaystyle \frac{1}{2}\frac{m}{s^2}$, with the result that its position at time t is $\displaystyle s(t) = s_{0} +v_{0} -\frac{1}{4}(t^2)$

show that $\displaystyle s_{0}$ is the train's postion when the brakes are applied and $\displaystyle v_{0}$ is its velocity at that time. Find the stopping distances if $\displaystyle v_{0}$ is $\displaystyle 20 \frac {m}{s}$

$\displaystyle s_{0} +v_{0}*(0) -\frac{1}{4}(0^2) = 0$ gives $\displaystyle s_{0} =0$

and

$\displaystyle v_{0} -\frac{1}{2}(0) = 0$ gives $\displaystyle v_{0} =0$

and solve s(t) to get the distance using $\displaystyle v_{0} = 20$

$\displaystyle 20 -\frac{1}{2}t\Rightarrow t =40$

$\displaystyle s(t) = 0 + 20(40) -\frac {1}{4}(40^2) = 400$

thus the answer is 400 metres.

ps. teaching myself LaTex