Physics, asked by AdvikaRai8673, 9 months ago

Joseph jogs from one end to the other end B of a straight 300 m road in 2 min 30 sec and then turns around jogs 100 m back to point C in another 1 min what are Joseph's average speed and velocities in jogging (a)from A to B and (b) from to C?

Answers

Answered by kiranshukla84626
8

Answer:

(a) Given,

Total distance from A to B = 300 m

Total time taken from A to B = 2 minutes 30 seconds

= 2 × 60 + 30

= 120 + 30

= 150 s

Average speed (from A to B) === 2.0 m/s (i)

The displacement and the time were taken of Joseph is the same in going from A to B i.e., 300m and 170 s respectively.

Therefore, Average velocity (from A to B) == = 2.0 m/s (ii)

So, from (i) and (ii) it is clear that the average speed and average velocity of Joseph during his jogging are the same.

(b) Given,

Total distance from A to C = 300 + 100= 400 m

Total time = 2 minutes 30 seconds + 1 minutes

= 150 s + 60 s (As we Know 1 Minute= 60 seconds)

= 210 s

Average Speed (from A to C) = = = 1.90 m/s (iii)

Now the average velocity of Joseph from A to C:

Displacement = 300 – 100= 200 m

Total time taken is the same as that of the time taken from A to C i.e., 210 s (Calculated Above)

Average velocity (from A to C) =

From (iii) and (iv) it is clear that the average speed of Joseph is different from his average velocity.

Explanation:

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