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

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

$a = \sqrt{a_t^2 + a_r^2}$

where;

$a_t$ is the tangential acceleration

$a_r$ is radial acceleration

$a_t = \frac{v-u}{t} \\a_t = \frac{15.556-23.889}{18} \\\\a_t = -0.463 \ m/s^2 \\\\a_t = 0.463 \ m/s^2 \ \ (inward)$

$a_r = \frac{v^2}{r} \\\\a_r = \frac{15.556^2}{160} \\\\a_r = 1.512 \ m/s^2$

$a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{(-0.463)^2+(1.512)^2} \\\\a = \sqrt{2.5005} \\\\a = 1.581 \ m/s^2$

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