A train is traveling south at 31.2 m/s when the brakes are applied. It slows down with constant acceleration to a speed of 6.00 m/s in a time of 9.00 s.What is the acceleration of the train during the 9.00-s interval? If the acceleration is toward north, enter a positive value. If the acceleration is toward south, enter a negative value.How far does the train travel during the 9.00 s? If the displacement is toward north, enter a positive value. If the displacement is toward south, enter a negative value.

The initial velocity [tex]u[/tex] of the train (before braking) will be negative since the the train was travelling south. Thus: [tex]u = (-31.2)\; {\rm m\cdot s^{-1}}[/tex].

The train continues southwards after braking. Therefore, the final velocity [tex]v[/tex] of the train (after braking) will also be negative: [tex]v = (-6.00)\; {\rm m \cdot s^{-1}}[/tex].

Find the change in the velocity of the train by subtracting the initial value from the final value:

[tex]\begin{aligned}\Delta v &= v - u \\ &= (-6.00)\; {\rm m\cdot s^{-1} - (-31.2)\; {\rm m\cdot s^{-1} \\ &= 25.2\; {\rm m\cdot s^{-1}} \end{aligned}[/tex].

Divide change in velocity by the time required [tex]t = 9.00\; {\rm s}[/tex] to find the (constant) acceleration [tex]a[/tex] of the train:

[tex]\begin{aligned} a &= \frac{\Delta v}{t} \\ &= \frac{25.2\; {\rm m\cdot s^{-1}}}{9.00\; {\rm s}} \\ &= 2.80\; {\rm m\cdot s^{-2}}\end{aligned}[/tex].

Note that acceleration is positive since the velocity is becoming less negative. The velocity component of the train towards the south has been reduced.

Apply the SUVAT equation [tex](v^{2} - u^{2}) = 2\, a\, x[/tex] to find the displacement [tex]x[/tex] of the train:

[tex]\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a} \\ &= \frac{(-6.00)^{2} - (-31.2)^{2}}{2 \times 2.80}\; {\rm m} \\ &\approx (-167)\; {\rm m} \end{aligned}[/tex].

Note that displacement is negative since the train keeps travelling south. The position of the train after braking is to the south of where the train was before braking.