High School

Clark is running on a straight road. He runs 500 meters north in 2 minutes and 10 seconds, then turns back and runs 200 meters south in 1 minute. Calculate his average speed and the magnitude of average velocity during the first 2 minutes and 10 seconds.

Answer :

Final answer:

Clark's average speed during his initial run is approximately 3.85 m/s and his average velocity during this time is 3.85 m/s towards the north.

Explanation:

Let's first understand the difference between speed and velocity. Speed is simply the rate at which an object covers distance. It is a scalar quantity as it doesn't take into account direction. Velocity, on the other hand, is a vector quantity. It involves both magnitude (speed) and direction.

Now coming to Clarks' case, he covers 500m in 2 minutes and 10 seconds (which is equivalent to 130 seconds). His speed during this phase would be the total distance divided by total time which gives approximately 3.85 m/s.

Now, as for the average velocity during the first 2 minutes 10 seconds, since he is running straight and not returning, the magnitude of displacement is equal to the distance, so it is also 3.85 m/s north.

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