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

when insulating materials rub against each other, they may become electrically charged . electrons , which are negatively charged, may be 'rubbed off' one material and on to the other. the material that gains electrons becomes negatively charged. the material that loses electrons is left with a positive charge.

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