Hello, Jeremy,
I don't think I've done this correctly, but take a look and see if you can make the approach work. If we plot the bus transit using km and with an orientation that the positive x is East and the negative y is South. We can calculate the km travelled for each of the three segments. I do South as an example:
(20m/s)*(60s/min)*(3min) = 3600m, or 3.6km due south. That would be a point (0, - 3.6) on our graph. 3.6 km due south.
The bus then turns right. I'm thinking I'm in the bus, so right to me means he turns left on the graph, towards the West. We can calculate a distance from the new speed and time. I get 4.6km due West. That would be a point that has the same y value, -3.6, but an x that subtracts 4.6 from the startying opint of 0. We we'll be at (-4.5, -3.6).
Then the bus heads northwest. We'll assume a 45 degree angle going to the left and up. The distance travelled is 5.4 km. We need the distances for the x and y components. If we draw a line from the point where the bus turned northwest, it will travel for 5.4 km, above the x axis. The 5.4km is the hypothenuse of an isoceles triangle. The two sides make up the x and y components. x the distance for x will equal the distance for y, since the triangle is isoceles. Y=Therefore, we can say fior this triangle, the two sides can be related to the hypothenuse:
x2 + y2 = (5.4)2:
I get x and y both = 3.82 km
Take the last point (-4.5, -3.6) and subtract 3.8 from the x value (we're moving left) and add 3.8 to the y value (heading up, north. That gives us the last point, (-8.32, 0.218).
These see like odd numbers, so I won't calculate any futher. But the final displacement is the distance between (0,0) and (-8.32, 0.218). This can be calculated using pythagoras' theorm. Use trig to find the angle.
Bob
