### 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