A train slows down as it rounds a sharp horizontal turn, going from 111.9 km/h t

Question

A train slows down as it rounds a sharp horizontal turn, going from 111.9 km/h to 54.6 km/h in the 15.6 s it takes to round the bend. The radius of the curve is 132.1 m. Compute the magnitude of the acceleration at the moment the train speed reaches 54.6 km/h. Assume the train continues to slow down at this time at the same rate. Give your answer in units of m/s

An explorer wishes to cross a river that is 1.8 km wide and that flows with a velocity of 5.9 km/h parallel to its banks. The explorer uses a small power boat that moves with a maximum speed of 7.1 km/h with respect to the water. If the explorer needs to cross directly to the opposite bank, what is the minimum time, in hours, needed to cross?

Explanation / Answer

At = dV/dt = (54.6 - 111.9)/(3.6*15.6) = -1.020 m/s² (the 3.6 is to change km/h to m/s)

Ar = V²/R = (54.6/3.6)²/132.1 = 1.741 m/s²

M = (At²+Ar²)

=(-1.020 m/s² )^2+(1.741)^2

M = 2.017 m/s^2

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

vr = 5.9km/hr, vb = vel of boat in still water, vb,r = 7.1 km/hr , width = 1.8 km

Using pythagoras th, we get vb = sqrt(7.12 - 5.92) = 3.95 km/hr

Tmin = width/vb = 1.8/3.95 = 0.45 hr

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.