# Video: Finding Position Vectors and Displacements in Two Dimensions in a Real-World Context

A car is 2.0 km west of a traffic light at the instant 𝑡 = 0 min and 5.0 km east of the same traffic light at 𝑡 = 6.0 min. Assume the origin of the coordinate system is the traffic light and the positive 𝑥-direction corresponds to eastward. What is the car’s position vector at 𝑡 = 0 min? What is the car’s position vector at 𝑡 = 6.0 min? What is the magnitude of the car’s displacement between 𝑡 = 0 min and 𝑡 = 6.0 min? What is the direction of the car’s displacement between 𝑡 = 0 min and 𝑡 = 6.0 min?

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### Video Transcript

A car is 2.0 kilometres west of a traffic light at the instant 𝑡 equals zero minutes and 5.0 kilometres east of the same traffic light at 𝑡 equals 6.0 minutes. Assume the origin of the coordinate system is the traffic light and the positive 𝑥-direction corresponds to eastward. What is the car’s position vector at 𝑡 equals zero minutes? What is the car’s position vector at 𝑡 equals 6.0 minutes? What is the magnitude of the car’s displacement between 𝑡 equals zero minutes and 𝑡 equals 6.0 minutes? What is the direction of the car’s displacement between 𝑡 equals zero minutes and 𝑡 equals 6.0 minutes?

Let’s begin our solution by drawing out a sketch of the car’s position over time. If the position of the traffic light, which is the origin, is marked by the 𝑥, then at time 𝑡 equals zero the car is at the position of negative 2.00 kilometres west of the origin, and at time 𝑡 equals 6.0 minutes, the car is at a position of positive 5.00 kilometres from the traffic light.

We want to know the position of the car at 𝑡 equals 0 minutes, the position of the car at 𝑡 equals 6.0 minutes, the car’s displacement between times 𝑡 equals zero minutes and 𝑡 equals 6.0 minutes, which we’ll call 𝑑. And finally, we want to know the direction that the car travelled over this time.

Regarding the position of the car at time 𝑡 equals zero minutes, when we look at our diagram, we see that its position is negative two units in the 𝑥-direction. On our chart, this position is negative 2.0 𝑖. The position of the car at time 𝑡 equals 6.0 minutes is positive five units along the 𝑥-axis, in other words, 5.0 𝑖. And this position like the position at time 𝑡 equals zero minutes is measured in units of kilometres.

When we solve for the displacement of the car over this time, that displacement is equal to its final position minus its initial position or 5.00 kilometres minus 2.00 kilometres, giving us a displacement to two significant figures of 7.0 kilometres.

And finally, we want to solve for the direction that the car travelled in. We’re told that it started two kilometres west of the traffic light and ended up five kilometres east of the traffic light. That means that east is in the direction of the positive 𝑥-axis. Since the car moved that way, it was moving east.