PHYSICS PROBLEM ON ONE DIMENSIONAL MOTION
Yesterday i found this interesting problem . Have fun in doing this.
On April 15 an airplane takes off at 4 : 40 p.m from Belem. Brazil bound for Villamil, Ecuador (in the Galapagos). The plane lands at 8 : 40 p.m. Villamil local time. The sun sets at 6 : 15 p.m. in Belem (local time), and 7 : 06 p.m. in Villamil (local time). At what time during the flight do the airplane passengers see the sun set?
On April 15 an airplane takes off at 4 : 40 p.m from Belem. Brazil bound for Villamil, Ecuador (in the Galapagos). The plane lands at 8 : 40 p.m. Villamil local time. The sun sets at 6 : 15 p.m. in Belem (local time), and 7 : 06 p.m. in Villamil (local time). At what time during the flight do the airplane passengers see the sun set?
what's the solution??
ReplyDeletelet the distance frm belem to
ReplyDeletevilamil is x km
nd the distance frm belem to
the place where the passenger
sees sunset is y km
sunsets inbelem at 6.15 nd in vilamil is 7.06
hence the time difference
between the two places is 7.06
- 6.15 = 0.51 hrs= 51 min
consider time according to
belem
the plane reaches vilamil at
8.40 local time..............
the time
in belem when the plane
reaches vilamil is 8.40 - 51 min =7.49. pm
therefore time taken for the
journey is 7.49 - 4.40= 3.09 hrs
let v (kmph) be the velocity
then x = v(3.09) km-----------1
the time taken for the sun set
frm begining of the journey is
6.15 - 4.40 =1.75 hrs then y = v (1.75) km-----------2
for x distance time variation is
0.51 hrs
for y distance time variation
is (T) T =0.51
dividing eqn 2 by
eqn 1
we get y/x = 1.75/3.09 therefore the time variance
btn the place where the
passebger sees the sunst nd
belem is
T = 0.51 X1.75/3.09=
0.28 hrs =
28 min
therefore the time at the place wer passenger sees
sunset is =
sunset at belem
+time difference btn belem nd
that place (T)
= 6.15 hrs+28 min =6.43 p.m
by prabhat kumar singh