A kayaker needs to paddle north across a 85 m wide harbor. The tide is going out, creating a tidal current that flows to the east at 1.6 m/s. The kayaker can paddle with a speed of 2.8 m/s.
(a) In which direction should he paddle in order to travel straight across the harbor?
To determine the direction the kayaker should paddle to travel straight across the harbor, we need to consider the relative velocities created by the tidal current and the kayaker's own paddling speed.
First, let's visualize the situation. We have a kayaker looking to travel north across an 85 m wide harbor. The tidal current is flowing to the east at a speed of 1.6 m/s, and the kayaker can paddle at a speed of 2.8 m/s.
To travel straight across the harbor, the kayaker needs to counteract the effect of the eastward tidal current and ensure their net velocity vector is directly north.
We can break down the velocities into their components along the north-south (y-axis) and east-west (x-axis) directions. Let's assume north is positive and east is positive.
The kayaker's paddling velocity can be split into two components: one along the y-axis (north) and one along the x-axis (east). Since the kayaker wants to travel straight across the harbor, the northward component of the kayaker's velocity needs to be equal to the eastward component of the tidal current velocity.
The eastward component of the kayaker's velocity is given by:
Eastward component of kayaker's velocity = kayaker's velocity * cosine (angle)
Eastward component of kayaker's velocity = 2.8 m/s * cosine(0°) (since the kayaker is paddling straight across)
The cosine of 0° is equal to 1, so the eastward component of the kayaker's velocity is 2.8 m/s.
Therefore, to counteract the eastward velocity caused by the tidal current, the kayaker needs to paddle with an eastward velocity of 2.8 m/s.
In conclusion, the kayaker should paddle eastward at a speed of 2.8 m/s to counteract the tidal current and travel straight across the harbor.
85/2.8=30.36 secs. to travel straight across. in the time he would have end up (30.36 x 1.6) = 48.57 m. downstream so now you can sketch a right triangle, 85 m. north/south, south 48.57 m. "upstream" on the far bank.
(a) angle to row = 48.57/85 = 0.57 deg, west of north.
(b) distance he actually paddles = sqrt. (85^2 + 48.57^2), 98 meters
time 98/2.8 = 35 secs