Question

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➜ SPAM AND IRREVALENT ANSWERS WILL BE REPORTED
➜ For SPAMMING, Answer will be reported
➜ Physics, class 9 chapter 8- Motion

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 1

Answer:

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Answer 2

Answer:-

$\textbf{(a) For Motion form A to B :}$

$★ \textbf{Distance Covered = 300m}$

$★ \textbf{Displacement = 300m}$

$★\text{Time Taken = 2 minutes 30 seconds}$

$= 2 \times 60 + 30 = 150s$

$\textbf{Average speed} = \frac{\text{Distance Covered } }{\text{Time Taken} }$

$⇒ \frac{300m}{150s} = 2 {ms}^{ - 1}$

$\textbf{Average Vilocity} = \frac{\text{Displacement } }{\text{Time Taken} }$

$⇒ \frac{300m}{150s} = 2 {ms}^{ - 1}$

$\textbf{(b) For motion from A to B to C:}$

$★ \textbf{Distance Covered = 300 + } 100 = 400 \textbf{m}$

$★ \textbf{Displacement = AB -CB}$

$= 300 - 10 = 200m$

$⇒\text{Time Taken = 150+60=210 s}$

$\textbf{Average speed} = \frac{\text{Distance Covered } }{\text{Time Taken} }$

$⇒ \frac{400m}{210} = 1.90 {ms}^{ - 1}$

$\textbf{Average Vilocity} = \frac{\text{Displacement } }{\text{Time Taken} }$

$⇒ \frac{200m}{210s} = 0.952 {ms}^{ - 1}$

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