### Chapter - 1 Motion

Q
##### Motion

Question:

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?

(a) From end A to end B

Distance covered by Joseph while jogging from A to B = 300 m

Time taken to cover that distance = 2 min 30 seconds = 150 s

Average speed $= \frac{Total\ distance\ covered}{Total\ time\ taken}$

Total distance covered = 300 m

Total time taken = 170 s

Average speed $= \frac{300}{150} =2\ m/s$

Average velocity $= \frac{Displacement}{Time\ interval}$

Displacement = shortest distance between A and B = 300 m

Time interval = 150 s

Average velocity $= \frac{300}{150} =2\ m/s$

The average speed and average velocity of Joseph for jogging from

A to B are the same and equal to 2 m/s

(b) From end A to end C

Average speed $= \frac{Total\ distance\ covered}{Total\ time\ taken}$

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

Total time taken = Time taken to travel from A to B + Time taken to travel from B to C = 150 + 60 = 210 s

Average speed$= \frac{400}{210} =1.90\ m/s$

Average velocity $= \frac{Displacement}{Time\ interval}$

Displacement from A to C = AC = AB - BC = 300 - 100 = 200 m

Time interval = Time taken to travel from A to B + Time taken to travel from B to C

Time interval = 150 s + 60 s = 210 s

Average velocity $= \frac{200}{210} =0.95\ m/s$

The average speed of Joseph from A to C is 1.90 m/s and his average velocity is 0.95 m/s.

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