A cyclist moves a distance of 4km toward east and then a dostance of 3km towards north. How far is he from the starting point?
Correct answer: D. 5km
- A. 3km
- B. 9km
- C. 7km
- D. 5km
Explanation
We have to find the displacement of the cyclist from the initial position. He moves 4km east (to the right) and 3km north (upwards). Since both directions are at 90 degrees to each other, a right triangle is formed (see diagram below). Using the Pythagorean theorem, you can find the third side by x = sqrt(42 + 32) = 5. Hence, the answer is D.
This question appeared on the ETEA MDCAT 2005 paper, which you can sit online with every answer explained.
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About Displacement and Velocity
Displacement is the directed change in position, while velocity is the rate of change of displacement and includes both magnitude and direction. The topic connects position-time and velocity-time graphs with average and instantaneous values, and distinguishes displacement from distance and velocity from speed.
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