Science, asked by sarthak6663, 9 months ago

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 speed and velocities in jogging:
a. From A to B ?
b. From A to C?​

Answers

Answered by PixieAngel
16

\huge\bold\blue{Answer -}

  • Refer the attachment for Answer (a).

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Answer (B)

Total distance covered from A to C= ( AB + BC)

=(300 + 100)m

= 400m

Total time taken from A to C=Time taken for AB + Time taken for BC

= (2 minutes 30 seconds + 1 minute)

= (150 + 60)seconds

=210 seconds

\tt\ {average \: speed \: from \: A \: to \: C}

\tt\ {=  \frac{total \: distance}{total \: time}  =   \frac{400m}{210s}  =  \frac{40}{21}  = 1.90m/s}

___ ___ ___ ___ ___ ___....

Total displacement (s) from A to C= AB - BC

= (300 - 100)m

= 200 m

Time taken for displacement from A to C= 210s

\tt\ {average \: velocity \: from \: A \: to \: C }

\tt\ {=  \frac{total\:displacement(s)}{total\:time(t)}  = \frac{200m}{210s}  =  \frac{20}{21}  = 0.95m/s}

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Answered by lucky4613
4

Answer:

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