Physics, asked by roanit333, 9 months ago

Joseph jogs from one end A to the other end B of a straight 300m road in 2 minutes 30 second and then turns around and jogs 100m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging from
a) A to B
b) A to C?

Answers

Answered by Lilyrose0111
17

Answer:

Velocity = dispacement / time

Speed = distance / time

a) when he jogs from A to B on a straight road,

displacement = distance = 300m

time = 2 minutes 30 seconds = 150 s

velocity = 300/150 = 2 m/s

speed = 300/150 = 2m/s

b)when he jogs from A to B and turns back to C,

displacement = 300-100 = 200m

distance = 300+100 = 400m

time = 3 minute 30 second = 210 s

velocity = 200/210 = 20/21 m/s

speed = 400/210 = 40/21 m/s

Answered by shahilkashyap858
11

Answer:

Hello mate!!

Here is your answer.!

⬇️

(a) Distance covered A➡️B = 300m; S=300m

★ Time taken = 2 min 50 sec

=( 2×60 +50 )

= 120+50

= 170 seconds .

★Average speed = Total distance/Total time

= 300 m / 170 s

= 30 / 17

= 1.76 ms-¹.

★Average velocity = Displacement / Total time

= 300 m / 170 s

= 1.76 ms-¹.

(b) From A➡️C

★ Total distance = A to B + B to C

= 300 +100

= 400 m

★Total time = 170+60

= 230 seconds.

★Average speed = 400 / 230

= 40 / 23

= 1.74 ms-¹.

★Average velocity = 200 / 230

= 20 / 30

= 0.87 m/s

♥️I hope it's helpful for you.

Mark as brainliest answer.♥️

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