a bird sitting on a Tree top at a height of 10 m from the ground piece of ostrich flying on the ground to build a nest on the build top it starts at 7 a.m. and ends at 7 a.m during this interval it makes 5 trips up and down find the distance and displacement and average speed
Answers
In triangle OAC tan 45=AC/OA=1
AC=OA=80
IN TRIANGLE ODB
tan 30=BD/OB=1/√ 3
OB=BD√ 3-AC√ 3(since ac=bd as bird remains constant height)
AB=OB-OA=AC√ 3-AC=AC(√ 3-1)=80(√ 3-1)
NOW FOR 2 SEC
80(√ 3-1)/2=40(√ 3-1)
=29.2M/SEC
hope u like
Answer with explanation:-
Total distance covered = 10m*2*5=100m.
Time taken = (07:10 - 07:00) minutes = 10 minutes=10*60 seconds=600seconds.
Formula of Avg. Speed = total distance/total time
By the formula,
the average speed = 100m/600s = 1/6m s^-1
Here, displacement=0.
It is because displacement is a vector quantity (quantity which has both magnitude and direction).As it returned to same place the direction did not changed at end. As there is no change in direction there is no displacement.
Note : Some of the readers may have problem in understanding why I multiplied 2 during calculating distance. It is because according to question the the bird made 5 trips up and down in 10 minutes. Here 1 trip = 10m(up)+10 m(down). as 1 trip results to 20 m I have written this as 10m*2 while calculating distance.