Question Video: Determining a Net Speed and Direction from Velocity Vector Components

The velocity vector of a polar bear is 𝑣 = (−18.0𝑖 − 13.0𝑗) km/h. Consider 𝑖 and 𝑗 to correspond to east and north, respectively. What is the speed of the polar bear? In what geographic direction is the bear heading?

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Video Transcript

The velocity vector of a polar bear is 𝑣 equals negative 18.0𝑖 minus 13.0𝑗 kilometers per hour. Consider 𝑖 and 𝑗 to correspond to east and north, respectively. What is the speed of the polar bear? In what geographic direction is the bear heading?

Let’s start off by drawing a sketch of this velocity vector. In our coordinate setup, we’re told that the northern and eastern directions correspond to positive 𝑗 and positive 𝑖 hat displacement, respectively. When we plot out the 𝑖- and 𝑗-components of our polar bear’s velocity vector, we see it points away from the north and away from the east.

We’re given the polar bear’s velocity vector, but we want to solve for the polar bear’s speed. These two terms are related by the relationship which says that speed is equal to the magnitude of velocity.

Since the 𝑖- and 𝑗-components of our velocity vector together form a right triangle with a hypotenuse of the velocity, we can say that the magnitude of velocity, the speed, is equal to the square root of each one of the components squared and then added together. to three significant figures, this is 22.2 kilometers per hour. That’s a pretty rapid speed, and I wouldn’t want to be near this polar bear if it was hungry.

Next, we want to solve for the direction of the polar bear’s travel. Considering our diagram, if we look at the right triangle that’s formed by the polar bear’s velocity vector as well as the 𝑥- and 𝑦-components of this vector, if we call the angle on the top right-hand corner 𝜃, we see we can solve for this angle using trigonometric identities.

In particular, we can say that the tangent of the angle 𝜃 is equal to the 𝑦-component of our polar bear’s motion, negative 13.0 kilometers per hour, divided by the 𝑥-component of that motion. If we take the inverse tangent of both sides of the equation, we now have an expression with 𝜃 by itself. When we calculate 𝜃, we find it’s 35.8 degrees. Using compass directions, that’s 35.8 degrees south of west. That’s the geographic direction of the polar bear’s travel.

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