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a kayaker can paddle her kayak at a steady 2.5 meters/sec in still water.

She wishes to cross a river that is 2.0 km wide and has a current of 1.5 meters/sec. If the kayaker aims her craft straight across the river, the current will carry her downstream as she paddles across. What will be her actual velocity (magnitude and direction angle) as she crosses? how long will it take her to cross the river
if she aims the kayak somewhat upstream, she can actually travel straight across the river. In what direction must she aim: What is her actual speed across the river for this situation, and how long will it take her to cross? 

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Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
5.0 5.0 (3 lesson ratings) (3)
In part 1, we have two perpendicular velocities due to the kayaker and the river: 2.5 m/s across the river and 1.5 m/s downstream. The magnitude of the resultant velocity follows from the Pythagorean theorem:
v= √(1.5²+2.5²) = 2.9 m/s.
The angle of the resultant velocity (relative to the shoreline) follows from the definition of the tangent function (opposite/adjacent):
θ= tan-1(2.5/1.5) = 59°
To find the time it takes her to cross the river at this angle, we first need the distance d she has to cover, which follows from the definition of the sine function (opposite/hypotenuse):
sinθ=2.0 km/d,
d=2.0km/sin(59°) = 2.33 km =2330 m
Then the time follows from the definition of velocity,
v=d/t,  t=d/v = 2330m/2.92m/s = 800 s.
(Notice that we keep 3 significant digits in the calculations, but round the final answer to 2 significant digits.)
In Part II, we want the resultant velocity to be across the river. This means the river's 1.5 m/s downstream velocity must be canceled by the boat's upstream velocity component. This upstream component follows from the definition of cosine (adjacent/hypotenuse):
where θ is the angle of the boat with respect to the shoreline (upstream). Therefore,
Her actual speed across the river is her boat's other velocity component, which follows from the definition of sine:
vacross=2.5m/s sin(53.1) =2.0 m/s.
At this speed, it will take her
t=2000 m/2.0 m/s =1000 s
to cross the river.
Note: the shortest path is not always the fastest!


Wish we could draw pictures on here! It would look somewhat like this:
        \ 2.5 (her still velocity)        
         \                              1.5 (river, downstream)
<     θ \                             ____>
  1.5 (her upstream component)
Draw a right triangle whose hypotenuse is her velocity in still water. The adjacent side must be her velocity component that's canceled by the river's velocity. The opposite side is her actual speed across the river, which we are asked to find. If you are familiar with the tip-to-tail rule, you can also do it this way: put the river's velocity arrow at the end of her velocity arrow such that the two arrows give you one arrow straight up. Again, her velocity is the hypotenuse in that triangle.
Brad M. | STEM Specialist plus Business, Accounting, Investment & EditingSTEM Specialist plus Business, Accountin...
4.9 4.9 (233 lesson ratings) (233)
Hey Chelsey -- if we aim straight across, we cross 2000m at 2.5m/s in 800s or 13hr 20min ... however, that 1.5m/s current during the 800s takes us 1200m downstream -- a good 2/3 mile off! Actual velocity estimate is 2.5 + 1.5/2 with a "10% discount" ~ 3m/s ... angle approx 1.5/3 * 60 deg ~ 30 deg downstream
Aiming upstream to negate the current should take longer than 800s ... form a 3-4-5 right triangle ... 1.5m/s upstream - 2.0m/s across - 2.5m/s paddle "diagonal" ... crossing now takes 2000/2.0 = 1000s or 16hr 40min -- an extra 3 hrs! Probably easier to aim straight and walk back upstream 2/3mi (15min walk). Aiming upstream sin ratio is 3/5 ...  multiply by 60 deg ~ 36 deg upstream aim. Regards :)


Always check your answer for plausibility: does it make sense to cross a 2000-m wide river in 13 or 16 hours at that speed?
Thank you AW -- good catch with the units ... a day on the river is better spent fishing than kayaking.
I'll leave my error publicly unedited -- much is learned making mistakes :)
Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...
4.8 4.8 (4 lesson ratings) (4)
Andrew W, is correct. Problem has to be read correctly and visualize it without diagram.
Vivian L. | Microsoft Word/Excel/Outlook, essay composition, math; I LOVE TO TEACHMicrosoft Word/Excel/Outlook, essay comp...
3.0 3.0 (1 lesson ratings) (1)
HI Chelsey;
I cannot solve the equation.
However, I can remind you that the width of the river is 2.0 km wide while the curent is 1.5 meters/sec.
You must convert.  The river is 2000 meters wide.
Sorry I cannot do more.