Answer:
The acceleration of the train is 1.581 m/s² inward.
Explanation:
Given;
initial velocity of the train, u = 86.0 km/h = 23.889 m/s
final velocity of the train, v = 56.0 km/h = 15.556 m/s
change in time, Δt = 18 s
The total acceleration of particles moving along a curved path is given as vector sum of the tangential acceleration and radial acceleration
[tex]a = \sqrt{a_t^2 + a_r^2}[/tex]
where;
[tex]a_t[/tex] is the tangential acceleration
[tex]a_r[/tex] is radial acceleration
[tex]a_t = \frac{v-u}{t} \\a_t = \frac{15.556-23.889}{18} \\\\a_t = -0.463 \ m/s^2 \\\\a_t = 0.463 \ m/s^2 \ \ (inward)[/tex]
[tex]a_r = \frac{v^2}{r} \\\\a_r = \frac{15.556^2}{160} \\\\a_r = 1.512 \ m/s^2[/tex]
[tex]a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{(-0.463)^2+(1.512)^2} \\\\a = \sqrt{2.5005} \\\\a = 1.581 \ m/s^2[/tex]
Therefore, the acceleration at the moment the train speed reaches 56.0 km/h is 1.581 m/s² inward.
