Question

joseph jogs from one end a to the other end b of a straight 300m road in 2 minutes 30 seconds and then turns around and jogs 100m back to point c in another 1 minute what are joseph average speed and velocities in jogging (a) from A to B and (b) fromA to C​

Answer 1

Explanation:

(A)

$\bf {Total \: distance \: covered \: from \: A \: to \: B-}$ = $\bf {300m}$

$\bf {Total \: time \: taken }$ = $\bf {2minutes 30 seconds}$

$\bf {= (120 + 30)s=150s}$

$\bf {Average \: speed \: from \: A \: to \: B-}$

$\bf { \frac{total \: distance}{total \: time \: taken} = \frac{300}{150} = 2 {ms}^{ - 1}}$

$\bf {Average \: velocity \: from \: A \: to \: B-}$

$\bf { \frac{total \: displacement (from \: A \: to \: B)}{total \: time}}$

$\bf {= \frac{300m}{150s} =2 {ms}^{ - 1} from \: A \: to \: B}$

________....._______....._______.....__________

(B)

$\bf {Total \: distance \: covered \: from \: A \: to \: C-}$ = $\bf {(AB + AC)}$ $\bf {= (300+100)m}$ = $\bf {400m}$

$\bf {Total \: time \: taken \: from \: A \: to \: C-}$ $\bf {(time \: taken \: for \: AB + time \: taken \: for \: BC)}$ $\bf { =2 minutes \: 30 seconds+ 1 minute}$

$\bf {= (150+60)s = 210 seconds }$

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$\bf {Average \: speed \: from \: A \: to \: C}$

$\bf = { \frac{total \: distance}{total \: time \: taken} = \frac{400}{210} = 1.90 {ms}^{ - 1}}$

$\bf {Total\: displacement (s) \: from \: A \: to \: C}$

$\bf { AB- BC= (300-100)m=200m}$

Time taken for displacement from A to C = 210 seconds.

$\bf {Average \: velocity \: from \: A \: to \: C-}$

$\bf {= \frac{total \: displacement(s)}{total \: time(t)}}$

$\bf {= \frac{200m}{210s}}$

$\bf {=0.95 {ms}^{ - 1}}$

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Answer 2

Answer:

Velocity = dispacement / time

Velocity = dispacement / timeSpeed = distance / time

Velocity = dispacement / timeSpeed = distance / timea) when he jogs from A to B on a straight road,

Velocity = dispacement / timeSpeed = distance / timea) when he jogs from A to B on a straight road,displacement = distance = 300m

Velocity = dispacement / timeSpeed = distance / timea) when he jogs from A to B on a straight road,displacement = distance = 300mtime = 2 minutes 30 seconds = 150 s

Velocity = dispacement / timeSpeed = distance / timea) when he jogs from A to B on a straight road,displacement = distance = 300mtime = 2 minutes 30 seconds = 150 svelocity = 300/150 = 2 m/s

Velocity = dispacement / timeSpeed = distance / timea) when he jogs from A to B on a straight road,displacement = distance = 300mtime = 2 minutes 30 seconds = 150 svelocity = 300/150 = 2 m/sspeed = 300/150 = 2m/s

Velocity = dispacement / timeSpeed = distance / timea) when he jogs from A to B on a straight road,displacement = distance = 300mtime = 2 minutes 30 seconds = 150 svelocity = 300/150 = 2 m/sspeed = 300/150 = 2m/sb)when he jogs from A to B and turns back to C,

Velocity = dispacement / timeSpeed = distance / timea) when he jogs from A to B on a straight road,displacement = distance = 300mtime = 2 minutes 30 seconds = 150 svelocity = 300/150 = 2 m/sspeed = 300/150 = 2m/sb)when he jogs from A to B and turns back to C,displacement = 300-100 = 200m

Velocity = dispacement / timeSpeed = distance / timea) when he jogs from A to B on a straight road,displacement = distance = 300mtime = 2 minutes 30 seconds = 150 svelocity = 300/150 = 2 m/sspeed = 300/150 = 2m/sb)when he jogs from A to B and turns back to C,displacement = 300-100 = 200mdistance = 300+100 = 400m

Velocity = dispacement / timeSpeed = distance / timea) when he jogs from A to B on a straight road,displacement = distance = 300mtime = 2 minutes 30 seconds = 150 svelocity = 300/150 = 2 m/sspeed = 300/150 = 2m/sb)when he jogs from A to B and turns back to C,displacement = 300-100 = 200mdistance = 300+100 = 400mtime = 3 minute 30 second = 210 s

Velocity = dispacement / timeSpeed = distance / timea) when he jogs from A to B on a straight road,displacement = distance = 300mtime = 2 minutes 30 seconds = 150 svelocity = 300/150 = 2 m/sspeed = 300/150 = 2m/sb)when he jogs from A to B and turns back to C,displacement = 300-100 = 200mdistance = 300+100 = 400mtime = 3 minute 30 second = 210 svelocity = 200/210 = 20/21 m/s

Velocity = dispacement / timeSpeed = distance / timea) when he jogs from A to B on a straight road,displacement = distance = 300mtime = 2 minutes 30 seconds = 150 svelocity = 300/150 = 2 m/sspeed = 300/150 = 2m/sb)when he jogs from A to B and turns back to C,displacement = 300-100 = 200mdistance = 300+100 = 400mtime = 3 minute 30 second = 210 svelocity = 200/210 = 20/21 m/sspeed = 400/210 = 40/21 m/s