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CBSE Class 9 Science Question 2 of 30

Motion — Question 2

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Question 2

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 and

(b) from A to C?

Answer

(a) From A to B

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 and (b) from A to C? NCERT Class 9 Science CBSE Solutions.

Given,

Distance covered from A to B = 300 m

Time taken to travel from A to B = 2 min and 30 sec = 60 + 60 + 30 = 150 sec

Displacement from A to B = 300 m

Average speed = total distance travelledtotal time taken\dfrac{\text{total distance travelled}}{\text{total time taken}}

Average velocity = total displacementtotal time taken\dfrac{\text{total displacement}}{\text{total time taken}}

Substituting we get,

Average speed = 300150\dfrac{300}{150} = 2 ms-1

Average velocity = 300150\dfrac{300}{150} = 2 ms-1

Hence, average speed from A to B is 2 ms-1 and average velocity = 2 ms-1

(b) From A to C

Distance covered from A to C = 300 m + 100 m = 400 m

Time taken from A to C = 2 min 30 sec + 1 min = 60 + 60 + 30 + 60 = 210 sec

Displacement from A to C = 300 m – 100 m = 200 m

Average speed = total distance travelledtotal time taken\dfrac{\text{total distance travelled}}{\text{total time taken}}

Average velocity = total displacementtotal time taken\dfrac{\text{total displacement}}{\text{total time taken}}

Substituting we get,

Average speed = 400210\dfrac{400}{210} = 1.9 ms-1

Average velocity = 200210\dfrac{200}{210} = 0.952 ms-1

Hence, average speed from A to C is 1.9 ms-1 and average velocity = 0.952 ms-1