Question

straight
Joseph jogs from one end A to the other end B of a
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:

so here he gas travelled from A to B = 300m

and B to C = 100 m

now we are supposed to find average speed and velocity

1) time took from A to B = 2min 30 sec = 150 sec

and distance travelled = 300m

average speed = 300 / 150 = 2m/s

now average velocity = 300 / 150 = 2m/s

2) so total distance travelled = 300 + 100 = 400m

and total time took from A to C = 1min + 150 sec

= 60 + 150= 210s

so average speed = 400/210

= 1.90 sec

now he has turned towards same path AB and stopped on point C so now his displacement will be = 300 - 100 = 200 m

and total time will be = 210s

so average velocity will be =

total displacement/ total time

200/210 = 0.95 m/s

Answer 2

AnsweR :-

From point A to B

Time taken, t = $2\:min\:30\:sec$

⠀⠀⠀⠀⠀⠀ ⠀⠀⠀$120+30 = 150\:s$

Distance,d = 300m

Average speed = $\frac{Total\:distance\: covered}{total\:Time\:taken}$

$\implies$ $\dfrac{300\:m}{150\:s}$

$\implies$$2.0ms^{-1}$

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Now joseph at point c then total time = 150 + 60 = 120s.

Now, total distance covered = AB + BC = 300+100=400m

$\therefore$ average speed = $\frac{Total\:distance\:covered}{Total\:time\:taken}$

$\implies$ $\dfrac{400}{210}$

$\implies$ $1.90ms^{-1}$

Now, the displacement is AC = AB-BC = 300-100=200m

$\therefore$ Average velocity = $\frac{Distance\: covered}{Time\:taken}$

$\implies$ $\dfrac{200}{210}$

$\implies$ $0.95ms^{-1}$

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