Question

A train slows down as it rounds a sharp horizontal turn, going from 86.0 km/h to 56.0 km/h in the 18.0 s that it takes to round the bend. The radius of the curve is 160 m. Compute the acceleration at the moment the train speed reaches 56.0 km/h. Assume the train continues to slow down at this time at the same rate.

Answer 1

The acceleration of the train is 1.581 m/s² inward.

Step-by-step 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

`a = √(a_t^2 + a_r^2)`

where;

`a_t` is the tangential acceleration

`a_r` is radial acceleration

`a_t = (v-u)/(t) \\a_t = (15.556-23.889)/(18) \\\\a_t = -0.463 \ m/s^2 \\\\a_t = 0.463 \ m/s^2 \ \ (inward)`

`a_r = (v^2)/(r) \\\\a_r = (15.556^2)/(160) \\\\a_r = 1.512 \ m/s^2`

`a = √(a_t^2 + a_r^2) \\\\a = √((-0.463)^2+(1.512)^2) \\\\a = √(2.5005) \\\\a = 1.581 \ m/s^2`

Therefore, the acceleration at the moment the train speed reaches 56.0 km/h is 1.581 m/s² inward.