Physics, asked by Anonymous, 6 months ago

Hey guys......need a hand..! QUESTION:- Joseph jogs from the one end A to the other end B of a straight 300m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another one minute. What is Joseph's average speed and velocity (a) from A to B (b) from A to C? No spam please❌❌​

Answers

Answered by trishabhuvi
4

Answer:

From pont A to B

Distance covered=300m

Displacement=300m

Time taken=2 min 30 sec=(2×60)+30=150 sec

Average speed=

Timetaken

Distance covered

=

150

300

=2 m/sec

Average velocity=

Timetaken

Displacement

=

150

300

=2 m/sec

From point A to C

Distance covered=300+100=400m

Displacement=300−100=200m

Time taken=3 min 30 sec=(3×60)+30=210 sec

Average speed=

Timetaken

Distance covered

=

210

400

=1.90 m/sec

Average velocity=

Timetaken

Displacement

=

210

200

=0.95 m/sec

Answered by Thatsomeone
12

Explanation:

 \tt For\:A\:to\:B \\ \\ \tt Distance = Displacement = 300 m \\ \tt Time = 2\:min\:30\:s = 120 + 30 = 150 s \\ \\ \tt We\:know\:that \\ \\ \tt Speed = \frac{distance}{time} \\ \\ \tt \longrightarrow Speed = \frac{300}{150} \\ \\  \longrightarrow \boxed{\bold{\red{\tt Speed = 2\:m\:{s}^{-1} }}} \\ \\ \tt Now\: Velocity = \frac{displacement}{time} \\ \\ \tt \longrightarrow Velocity = \frac{300}{150} \\ \\ \tt \longrightarrow \boxed{\bold{\red{ Velocity = 2 \:m\:{s}^{-1} }}} \\ \\ \tt For\:A\:to\:C \\ \\ \tt Distance = 300 + 100 = 400 m \\ \tt Displacement = 300 - 100 = 200 m \\ \tt Time = 150 s + 60 s = 210\:s \\ \\ \tt Average\:Speed = \frac{Total\: distance}{Total\:time} \\ \\ \tt \longrightarrow Speed = \frac{400}{210} \\ \\ \boxed{\bold{\red{\tt Speed = \frac{40}{21} \:m\:{s}^{-1} }}}\\ \\ \tt Velocity = \frac{displacement}{total\:time} \\ \\ \tt \longrightarrow Velocity = \frac{200}{210} \\ \\ \boxed{\bold{\red{\tt Velocity = \frac{20}{21}\:m\:{s}^{-1} }}}

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