Question

8) A train enters a curved horizontal section of the track at a speed of 100 km/h and slows down with constant deceleration to 50 km/h in 12 seconds. If the total horizontal acceleration of the train is 2 m/s2 when the train is 6 seconds into the curve, calculate the radius of curvature of the track for this instant.

Answer 1

the radius of curvature of the track for this instant is 266 m

Explanation:

Given that;

The Initial Velocity u = 100 km/h = 100 ×
`(5)/(18)` = 27.77 m/s

velocity of the train at t=12 s is;

`V_(t=12)` = 50 km/h = 50 ×
`(5)/(18)` = 13.89 m/s

now, we calculate the deceleration of the train

`V_(t=12)` = u + at

13.89 = 27.77 +
`a_(t)`12

`a_(t)` = (13.89 - 27.77) / 12

`a_(t)` = -13.88 / 12

`a_(t)` = - 1.1566 m/s²

Now, the velocity of the train at 6 seconds is;

`V_(t=6)` = u + at

`V_(t=6)` = 27.77 + ( - 1.1566 m/s²)6

`V_(t=6)` = 27.77 - 6.9396

`V_(t=6)` = 20.83 m/s

The acceleration at t=6 s is;

a = √[ (
`a_(t)` )² + (
`a_(n)`)²]

a = √[ (
`a_(t)` )² + (
`a_(n)`)²]

we substitute

2m/s² = √[ (- 1.15 )² + (
`a_(n)`)²]

4 = (- 1.1566 )² + (
`a_(n)`)²

4 = 1.3377 + (
`a_(n)`)²

(
`a_(n)`)² = 4 - 1.3377

(
`a_(n)`)² = 2.6623

`a_(n)` = √2.6623

`a_(n)` = 1.6316 m/s²

Now the radius of curve is;

a = V² / p

`p_(t=6)` = (
`V_(t=6)` )² /
`a_(n)`

`p_(t=6)` = ( 20.83 m/s )² / 1.6316 m/s²

`p_(t=6)` = 433.8889 / 1.6316

`p_(t=6)` = 265.9 m ≈ 266 m

Therefore; the radius of curvature of the track for this instant is 266 m