Two cars approach an intersection. Car A is traveling due west at 15 m/s, and car B is traveling due north at 20 m/s. What is the velocity of car A as seen by the driver in car B?

To find the velocity of car A as seen by the driver in car B, we need to find the relative velocity of car A with respect to car B.

Let's denote the velocity of car A as VA = 15 m/s (west) and the velocity of car B as VB = 20 m/s (north).

The relative velocity of car A with respect to car B (VA/B) can be found using vector addition:

VA/B = VA - VB

Since car A is traveling west and car B is traveling north, we need to consider the directions of the velocities. We can break down the velocities into their x and y components:

VA = -15 m/s (west) = -15i
VB = 20 m/s (north) = 20j

Now, we can find the relative velocity of car A with respect to car B:

VA/B = VA - VB
VA/B = -15i - 20j

Therefore, the velocity of car A as seen by the driver in car B is -15 m/s (west) - 20 m/s (north).

Let's use the relative velocity formula:

```
v_rel = v_obj - v_obs
```

where:

- v_rel is the velocity of the object relative to the observer
- v_obj is the velocity of the object
- v_obs is the velocity of the observer

In this case, the observer is the driver in car B and the object is car A.

We can break down the velocities of both cars into their x and y components:

```
v_Ax = -15 m/s (due west)
v_Ay = 0 m/s
v_Bx = 0 m/s
v_By = 20 m/s (due north)
```

The relative velocity of car A as seen by the driver in car B is then:

```
v_relx = v_Ax - v_Bx = -15 m/s
v_rely = v_Ay - v_By = -20 m/s
```

The magnitude of the relative velocity is:

```
|v_rel| = sqrt(v_relx^2 + v_rely^2) = sqrt((-15 m/s)^2 + (-20 m/s)^2) = 25 m/s
```

The direction of the relative velocity is:

```
theta = arctan(v_rely / v_relx) = arctan(-20 m/s / -15 m/s) = 53.13 degrees below the horizontal
```

Therefore, the velocity of car A as seen by the driver in car B is 25 m/s at 53.13 degrees below the horizontal.