Question

A jogger jogs from one end to the other of a straight 300 meter track (from point A to point B in 2.50 min and then turns around and jogs 100 m back toward the starting point to point C) in another 50 s. What is the jogger’s average speed?

Answer 1

Answer:

sorrry i don't understand

Answer 2

Time = 2 min 50s=(2×60)+50s=170s
50
s
=
(
2
×
60
)
+
50
s
=
170
s


(i) Average speed =Distance coveredTime=300170=1.76ms−1
=
Distance covered
Time
=
300
170
=
1.76
m
s
-
1

Average velocity =Displacement along ABTime
=
Displacement along AB
Time

=300170=1.76ms−1
=
300
170
=
1.76
m
s
-
1
along AB
(ii) When joseph turns around fron B to C to towards A.
Average speed =Distance coveredTime=400230=1.7ms−1
=
Distance covered
Time
=
400
230
=
1.7
m
s
-
1

Average velocity =Distance ACTime=200230
=
Distance AC
Time
=
200
230
along AC
0.87ms−1
0.87
m
s
-
1
along AC