Asked by Ryomasa
A train slows down as it rounds a sharp bend, slowing from 90.0km/h to 50.0km/h in the 15.0s that it takes to round the bend. The radius of the curve is 150m.
Compute the acceleration at the moment the train speed reaches 50.0km/h. Assume it continues to slow down at the same rate.
Compute the acceleration at the moment the train speed reaches 50.0km/h. Assume it continues to slow down at the same rate.
Answers
Answered by
drwls
The acceleration is the vector sum of a centripetal component and a tangential component (along the direction of motion). The two componets are perpendicular.
The centripetal component is
a1 = V^2/R
Make sure you convert the 50 km/hr to meters per second, to get the answer in m/s^2. 50,000m/3600 s = 13.89 m/s
The 40 km/h speed reduction is 11.11 m/s. Divide that by 15.0 s for the tangential acceleration in m/s^2.
The centripetal component is
a1 = V^2/R
Make sure you convert the 50 km/hr to meters per second, to get the answer in m/s^2. 50,000m/3600 s = 13.89 m/s
The 40 km/h speed reduction is 11.11 m/s. Divide that by 15.0 s for the tangential acceleration in m/s^2.
Answered by
Ryomasa
How did you find the speed reduction?
Answered by
drwls
The speed reeduction is 90 - 50 km/h. That is 40 km/h
Answered by
Ryomasa
Converted into m/s, I presume?
Answered by
Ryomasa
Converted to m/s, I presume?
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