Examcrackers 1001 Physics (5th ed.) Question #22 and #26

[DynamicEnigma]

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#22 A particle moves from position 1 to position 2 along a path C. It travels at a constant speed of 5pi m/s. At exactly half way through the trip its average vertical velocity is?

The answer is 10 m/s. I understand how to obtain the correct answer by dividing vertical displacement (5m) by time which is equal to distance / speed. In this case speed is given, but I don't know how they found the distance of 2pi5/4.

Similarly, #26 asks for the average acceleration of a particle moving from position 1 to position 2 along path C going a constant speed of 1 m/s.

The answer is 0.4/pi m/s^2. Acceleration = change in velocity / time. As with the last problem time is equal to distance / speed, and I am able to find all the variables besides distance. According to the answer key distance in this question is given as 2pi5/2.

How might I find the distance of 2pi5/4 in #22 and 2pi5/2 in #26?

Any help would be much appreciated! 🙂

 

[pgoat]

22. Path C is a half-circle, and the distance at half way of that is 1/4 the circumference of the entire circle. Circumference is 2 * radius * pi, so the distance we're looking for is (2 * 5 * pi) / 4. Distance = speed * time, so now we can solve for time and we end up with t = 1/2. Then vertical displacement / time = 10m.

26. Similar to the last problem, distance is the circumference of circle / 2 (since it's a half-circle).

 

[DynamicEnigma]

22. Path C is a half-circle, and the distance at half way of that is 1/4 the circumference of the entire circle. Circumference is 2 * radius * pi, so the distance we're looking for is (2 * 5 * pi) / 4. Distance = speed * time, so now we can solve for time and we end up with t = 1/2. Then vertical displacement / time = 10m.

26. Similar to the last problem, distance is the circumference of circle / 2 (since it's a half-circle).

Thanks for the reply! I don't know how I was missing that.