Question

a freight train moving at an initial speed of 40 m/s puts on its brakes, producing a deceleration of .50 m/s^2.
a. how long will it take the train to travel the next 100m?
b. at what speed will it be traveling at the end of 100m?

Answer 1

Again just use the equations of motion here as in the previous question below. You know that the acceleration is -0.5 m/s\(^2\). You know the original speed is 40 m/s. we want to find s, but we dont know teh final speed yet. So the equation \[s=ut+\frac{1}{2}at^2\] will suffice. plugging numbers in we get \[100=(40)t+\frac{1}{2}(-0.5)t^2\]rearranging we get \[0.25t^2-40t+100=0\]which is a qudratic equation with \(t\) as its subject. We can solve this through factorisation, or using the general solution for quadratic equation of form \(at^2+bt+c=0\) \[t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] substituting in numbers we get \[t=\frac{-(-40)\pm\sqrt{(-40)^2-4(0.25)(100)}}{2(0.25)}\]. There will be two answers 2.54 seconds and 157.46 seconds because of the square root sign. Which one to take though? This will be answered below by considering the final speed.

We can calculate the final speed using \[v^2=u^2+2as\] or \[v=\sqrt{u^2+2as}=\sqrt{40^2+2(-0.5)(100)}=38.73\]. There are of course again two answers because of the square root so the answer is +38.73 and -38.73 m/s.

So which one do we take as being correct? Well strictly speaking both are. The reason being is that we are not dealing with speed and distance but rather velocity and displacement both of which have direction. The negative velocity is a valid solution and represents the train slowing to stop, and then reversing. In that it will move past the 100 meter mark, carry on down the track, and then reverse to back to the 100 meter mark. Remember that if I go a distance of 250 meters forward and 150 meters backwards, it is the same displacement as just going 100 meters.

In this situation we see that after 2.54 seconds the train will pass the 100 metre point passing it at 38.73 m/s, but that if it maintained the same constant acceleration that after 157.46 seconds, it would pass the same point but going the other way at 38.73 m/s. To check this use the equation \[v=u+at\] for \(v=\pm38.73\) m/s to find the corresponding times (remembering that acceleration is negative). For \(v=+38.73\) m/s we get \(t = 2.54\) seconds and for \v=-38.73\) m/s we get \(t = 157.46\) seconds.