Question

(b) Fig. 1.1 shows two airports A and C. north SE с sea land क WE not to scale Fig. 1.1 An aircraft flies due north from A for a distance of 360 km (3.6 x 105 m) to point B. Its average speed between A and B is 170ms-1. At B the aircraft is forced to change course and flies due east for a distance of 100 km to arrive at C. (0) Calculate the time of the journey from A to B.​

Answer 1

The addition of vectors and the uniform motion allows to find the answers for the questions about distance and time are:

• The distance to go between airports A and C  is 373.6 10³ m

• The time to go from airport A to B is 2117 s

Vectors are quantities that have modulus and direction, so their addition must be done using vector algebra.

In this case the plane flies towards the North a distance of y = 360 10³ m at an average speed of v = 170 m / s, when arriving at airport B it turns towards the East and travels from x = 100 10³ m, until' it the distance reaches the airport C

Let's use the Pythagoras theorem to find the distance traveled

               R = Ra x² + y²

               R =   10³

               R = 373.6 10³ m

They indicate the average speed for which we can use the uniform motion ratio

               v = $\frac{\Delta y }{t}$

                t = $\frac{\Delta y}{v}$

They ask for the time in in from airport A to B, we calculate

                t = 360 10 ^ 3/170

                t = 2.117 10³ s

In conclusion we use the addition of vectors the uniform motion we can find the answer for the question of distance and time are:

• The distance to go between airports A and C B is 373.6 10³ m

• The time to go from airport A to B is 2117 s

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