Greg walks on a straight road from his home to a convenience store 3.0 km away with a speed of 6.0 km/h. On reaching the store he finds that the store is closed, he instantly turns around and walks back with a speed of 9.0 km/h. Round your final answers to 2 significant figures.

a. What is the magnitude of average velocity and average speed of Greg over the first 30 mins of his travel?

Al comment me over here!

This question was apprently selected from the "Sneaky Questions" category.

The store is 3 km from his home, and he walks there with a speed of 6 km/hr. So it takes him (3 km) / (6 km/hr) = 1/2 hour to get to the store.

That's 30 minutes. So the whole part - (a.) of the question refers to only that part of the trip, and we don't care what happens once he reaches the store.

a). Over the first 30 minutes of his travel, Greg walks 3.0 km on a straight road, and he ends up 3.0 km away from where he started.

Average speed = (distance/time) = (3.0 km) / (1/2 hour) = 6.0 km/hr

Average velocity = (displacement/time) = (3.0 km) / (1/2 hour) = 6.0 km/hr

There's probably some more questions in part - (b.) where you'd need to use Greg's return trip to find the answers, but johnaddy210 is only asking us for part - (a.).