The train on a section of the track with a radius of 700 m for 30 seconds of equal slowing down reduced

The train on a section of the track with a radius of 700 m for 30 seconds of equal slowing down reduced the speed 72 to 54 km / h. by how much did the full acceleration drop?

Given: R (radius of the site) = 700 m; t (time) = 30 s; the movement is equally slow; V0 (initial speed) = 72 km / h (20 m / s); V (final speed) = 54 km / h (15 m / s).

1) Initial moment of time: а1 = √ (аτ ^ 2 + аn ^ 2) = √ (((V – V0) / t) ^ 2 + (V0 ^ 2 / R) ^ 2) = √ (((15 – 20) / 30) ^ 2 + (202/700) ^ 2) = 0.595 m / s2.

2) End of deceleration: a2 = √ (aτ ^ 2 + an ^ 2) = √ (((V – V0) / t) ^ 2 + (V ^ 2 / R) ^ 2) = √ (((15 – 20 ) / 30) ^ 2 + (152/700) ^ 2) = 0.362 m / s2.

3) Decrease in total acceleration: Δa = a1 – a2 = 0.595 – 0.362 = 0.233 m / s2.