Joseph jogs from one end A to the other end b of a straight 300

1.	Joseph jogs from one end A to the other end b of a straight 300
Answer» ong>Explanation: (a) A to B Total distance (s) from A to B = 300M Time (t) = 2min 30s = (2 x 60) + 30 = 150s AVERAGE speed from A to B= s/t = 300m/150 = 2m/s Total displacement (s) from A to B = 300m Time (t) = 2min 30s = (2 x 60) + 30 = 150s Average VELOCITY From A to B = s/t = 300m/150 = 2m/s (b) A to C Total distance (s) from A to C = 300m + 100m = 400M Time (t) = 2min 30s + 1min = [(2 x 60) + 30] + [60] = 210s Average speed from A to C = s/t = 400m/210 = 1.90m/s Total displacement (s) from A to C = 200m Time (t) = 2min 30s + 1min = [(2 x 60) + 30] + [60] = 210s Average velocity from A to C = s/t = 200m/210 = 0.95m/s

Answer»

ong>Explanation:

(a) A to B

Total distance (s) from A to B = 300M

Time (t) = 2min 30s = (2 x 60) + 30 = 150s

AVERAGE speed from A to B= s/t = 300m/150 = 2m/s

Total displacement (s) from A to B = 300m

Time (t) = 2min 30s = (2 x 60) + 30 = 150s

Average VELOCITY From A to B = s/t = 300m/150 = 2m/s

(b) A to C

Total distance (s) from A to C = 300m + 100m = 400M

Time (t) = 2min 30s + 1min = [(2 x 60) + 30] + [60] = 210s

Average speed from A to C = s/t = 400m/210 = 1.90m/s

Total displacement (s) from A to C = 200m

Time (t) = 2min 30s + 1min = [(2 x 60) + 30] + [60] = 210s

Average velocity from A to C = s/t = 200m/210 = 0.95m/s

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