Answer:
[tex]-0.44 m/s^2[/tex]
Explanation:
First of all, we need to calculate the distance covered by the locomotive during the reaction time, which is
t = 0.46 s
During this time, the locomotive travels at
v = 13 m/s
And the motion is uniform, so the distance covered is
[tex]d_1 = vt = (13)(0.46)=6.0 m[/tex]
The locomotive was initially 200 m from the crossing, so the distance left to stop is now
[tex]d=200 - 6.0 = 194.0 m[/tex]
And now the locomotive has to slow down to a final velocity of [tex]v=0[/tex] in this distance. We can find the minimum deceleration needed by using the suvat equation:
[tex]v^2 - u^2 = 2ad[/tex]
where
v = 0 is the final velocity
u = 13 m/s is the initial velocity
a is the deceleration
d = 194.0 m is the distance to stop
Solving for a,
[tex]a=\frac{v^2-u^2}{2d}=\frac{0^2-13^2}{2(194)}=-0.44 m/s^2[/tex]
