asuma heads directly across a river swimming at 1.6 m/s relative to still water she arrives 8.40 downstream from the point directly across the river which is 80 metre Wide
1. what is the speed of the river current??
2. what is this swimmer's speed relative to the shore ??
in what direction should the summer head so as to arrive at a point directly opposite to to her starting point ???
Answers
Answer:
Here V
R
= velocity of river
V
M
= velocity of man in still water = 2m/s (given)
from figure it is clear that
sinθ=
V
M
V
R
, where θ=60 degree
⟹V
R
=
3
m/s
from figure it is evident that
V
M
cosθ×t=1000m
⟹t=1000 s
Since speed and distance are directly proportional, the ratio of the downstream distance to the width of the river is the same as the ratio of the current speed to the swimmer's speed.
x = vx
y vy
40 m = vcurrent
80 m 1.6 m/s
vcurrent = 0.8 m/s
Determining the resultant velocity is a simple application of Pythagorean theorem.
v2 = vx2 + vy2
v = √[(0.8 m/s)2 + (1.6 m/s)2]
v = 1.8 m/s
Direction angles are often best determined using the tangent function. This problem is no exception. The only thing open to discussion is our choice of angle. I suggest using the angle between the resultant velocity and the displacement vector that points directly across the river, but this is just my preference. Be sure to indicate that the resultant lies on a particular side of this vector for clarity.
tan θ = x = vx
y vy
tan θ = 40 m = 0.8 m/s
80 m 1.6 m/s
tan θ = 0.5
θ = 27° downstream
This is where it gets interesting. By now you should understood that time is the ratio of displacement to velocity. This is a vector problem, so direction matters. This is why we should probably use the words displacement and velocity instead of distance and speed. The only question is which distance and which speed should we use? The simple answer is pick the pair you like the best, just be sure they point in the same direction. It works along either of the component directions…
t = x = y
vx vy
t = 40 m = 80 m
0.8 m/s 1.6 m/s
t = 50 s
It also works along the resultant direction…
t = r
v
t = √[(40 m)2 + (80 m)2]
√[(0.8 m/s)2 + (1.6 m/s)2]
t = 50 s
The time it takes to cross a river by a swimmer swimming straight across is independent of the speed of the river. The only factors that matter are the speed of the swimmer and the width of the river. This swimmer will always cross the river in 50 s regardless of the speed of the river. 1 m/s, 10 m/s, 100 m/s, it doesn't matter. This example is a perfect illustration of an idea to be presented in the next section of this book. Motion in two dimensions can be thoroughly described with two independent one-dimensional equations. This idea is central to the field of analytical geometry.