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.

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

Explanation:

Given that;

The Initial Velocity u = 100 km/h = 100 × = 27.77 m/s

velocity of the train at t=12 s is;

= 50 km/h = 50 × = 13.89 m/s

now, we calculate the deceleration of the train

 = u + at

13.89 = 27.77 + 12

= (13.89 - 27.77) / 12

= -13.88 / 12

= - 1.1566 m/s²

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

 = u + at

 = 27.77 + ( - 1.1566 m/s²)6

 = 27.77 - 6.9396

 = 20.83 m/s

The acceleration at t=6 s is;

a = √[ ( )² + ()²]

a = √[ ( )² + ()²]

we substitute

2m/s² = √[ (- 1.15 )² + ()²]

4 = (- 1.1566 )² + ()²

4 = 1.3377 +  ()²

()² = 4 - 1.3377

()² = 2.6623

= √2.6623

 = 1.6316 m/s²

Now the radius of curve is;

a = V² / p

= ( )² /  

= ( 20.83 m/s )² /  1.6316 m/s²

= 433.8889 / 1.6316

= 265.9 m ≈ 266 m

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

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