A train slows down as it rounds a sharp horizontal turn, slowing from 97.4 km/h to 47.8 km/h in the 12.9 s that it takes to round the bend. The radius of the curve is 136 m. Compute the acceleration at the moment the train speed reaches 47.8 km/h. Assume that it continues to slow down at this time at the same rate.
FIND: Magnitude & Direction °
2007-10-25 22:24:33 · 1 answers · asked by Justin 3 in Science & Mathematics ➔ Physics
Tangential component
at=(V2-V1)/(t2-t1) since we need to convert km/h to m/s use multiplication factor 0.27778
at= (47.8 - 97.4)(0.27778) /12.9=
Centripetal component
ac=V^2/R
ac=(47.8 x 0.27778)^2/136
the total acceleration is
a=sqrt(at^2 + ac^2)
with an angle
A=arctan(ac/at) direction apposite to motion and towards the center of the curve.
2007-10-26 01:10:40 · answer #1 · answered by Edward 7 · 1⤊ 0⤋