# Skill Set 03.3

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Main Page > IB Physics Skill Sets > Skill Set 03.3

### 1. A boat points straight across a 68.0 m wide river and crosses it in 18.2 seconds. In doing this it is carried downstream 23.7 m. What the boat’s velocity with respect to the water?

All that this question is asking for is the velocity the boat must've gone in order to cross the 68 m river in 18.2 s. To solve for velocity, we divide the distance by the time.

(68 m)/(18.2 s) = 3.7362637 = 3.74 m/s

### 2. A boat points straight across a 68.0 m wide river and crosses it in 18.2 seconds. In doing this it is carried downstream 23.7 m. What is the speed of the current?

If the boat went straight across, the only downward velocity is that of the current. So in order to find out how fast the current was going to carry the boat 23.7 m down in 18.2 s, we solve for velocity the same way we did in #1.

(23.7 m)/(18.2 s) = 1.302197 = 1.30 m/s

### 3. A boat points straight across a 68.0 m wide river and crosses it in 18.2 seconds. In doing this it is carried downstream 23.7 m. What is the velocity (in angle magnitude notation) of the boat as it moves across the river?

Using the x velocity we got in #1 and the y velocity we got in #2, we can solve for angle magnitude by using inverse tangent and Pythagorean theorem.

a2 + b2 = c2
3.742 + 1.302 = c2
c = 3.96 m/s
tan-1(1.30/3.74) = 19.167o
3.96 m/s @ 19.2o

### 4. A boat has a velocity of 3.72 m/s in a river with a current of 1.31 m/s. The river is 47.2 m wide. If you point straight across the river, what time will it take you to cross?

Basically the same problem as #1. Divide to get the time instead of the velocity though.

(47.2 m)/(3.72 m/s) = 12.68817 = 12.7 s

### 5. Question

Solution goes here