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Nata [24]
3 years ago
8

A boat is moving in a river with a current that has speed vW with respect to the shore. The boat first moves downstream (i.e. in

the direction of the current) at a constant speed, vB , with respect to the water. The boat travels a distance D in a time tOut . The boat then changes direction to move upstream (i.e. against the direction of the current) at a constant speed, vB , with respect to the water, and returns to its original starting point (located a distance D from the turn-around point) in a time tIn .
1) What is tOut in terms of vW, vB, and D, as needed?
2) What is tIn in terms of vW, vB, and D, as needed?
3) Assuming D = 120 m, tIn = 170 s, and vW = 0.3 m/s, what is vB, the speed of the boat with respect to the water?
4) Once again, assuming D = 120 m, tIn = 170 s, and vW = 0.3 m/s, what is tOut, the time it takes the boat to move a distance D downstream?
Physics
1 answer:
slega [8]3 years ago
6 0

Answer:

Explanation:

Current  has speed vW with respect to the shore and boat has speed vB with respect to water or current so speed of boat  with respect to shore

vW + vB .

Distance travelled with respect to shore by boat = D

time ( tout ) = distance / speed with respect to shore

tOut = D / ( vW + vB )

When the boat travels upstream , its velocity with respect to shore

= ( vB - vW ) , vB must be higher .

tin = D /  ( vB - vW )

3 ) tin = D /  ( vB - vW )

170 = 120 / (vB - 0.3 )

(vB - 0.3 ) = 12 / 17 = .706

vB = 1.006 m / s

4 )

tOut = D / ( vW + vB )

= 120 / ( .3 + 1.006 )

= 92.26 s

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