A small motor boat is capable of maintaining a 8.6 m/s speed in still water. It
ID: 1963280 • Letter: A
Question
A small motor boat is capable of maintaining a 8.6 m/s speed in still water. It attempts to cross a long straight 0.5 km wide river that has a constant 3.6 m/s current.(a) If the bow of the boat is aimed directly across the river, what will be the crossing time?
s
(b) For the same conditions in part (a), how far downstream will the boat reach the opposite shore?
m
(c) If the bow of the boat is aimed upstream to compensate for the current so as to arrive at the opposite shore directly across from the departure point, what is the "crab angle" (ie. the angle between the bow's direction and the actual path of the boat.)?
degrees
(d) For the same conditions in part (c), what will be the crossing time?
Explanation / Answer
a) We know that he will be moving constantly across the water, so the the crossing time is dependent on his horizontal velocity, which is given by:
v = x / t so that:
t = x / v = (400 m)/(9.0 m/s) = 44.4 seconds
b) Since he is moving 44.4 seconds horizontally, we know during that time he is also drifting downstream, so that the distance of his drift is:
x = vt = (3.6 m/s)(44.4 s) = 160 m.
c) Since the boat moves at a velocity of 9.0 m/s straight ahead, we want to angle to boat so that the vertical component of its velocity cancels out with the velocity of the stream, so his velocity is purely horizontal:
sin = (3.6 m/s) / (9.0 m/s)
= sin-1(3.6 / 9.0) = 23.58o
d) Now his horizontal velocity has changed, although the same concept as from part (a) still applies. We can find his new horizontal velocity by:
cos(23.58) = v / 9.0
v = 9cos(23.58) = 8.25 m/s
So again:
t = x / v = (400 m) / (8.25 m/s) = 48.5 s
*Make sure all answers are in correct sig figs..... I didn't do them here, as your teacher may have some specific rule*
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