. Joseph jogs from one end A to the other end B of a straight 300 m road


in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging?

(a) From A to B

 (b) From A to C

 

Best Answer

Ans. Firstly, we have to draw a line segment to show the movement of Joseph during his jogging

we have to draw a segment

(a) Given,

 Total distance from A to B = 300 m

 Total time taken = 2 minutes 30 seconds

 = 2 × 60 + 30

 = 120 + 30

 = 150 s

Average speed (from A to B) = Total distance/Total time taken

= 300m/150s

The displacement and the time taken of Joseph is same ingoing from A to B i.e., 300m and 150 s respectively.

Therefore, Average velocity (from A to B) = Displacement/toatal time taken

= 300m/150s

 = 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 is same.

(b) Given,

 Total distance from A to C = 300 + 100

 = 400 m

 Total time = 2 minutes 20 seconds + 1 minutes

 = 150 s + 60 s

 = 210 s

Average Speed (from A to C) = Total distance/Total time taken

=400/210s

= 1.90 m/s (iii)

Now the average velocity of Joseph from A to C:

Displacement = 300 – 100

 = 200 m

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

Average velocity (from A to C) = Total distance/total time taken

= 200m/210s

= 0.95 m/s (iv)

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

 

 

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