In: Physics
In reaching her destination, a backpacker walks with an average velocity of 1.38 m/s, due west. This average velocity results because she hikes for 6.44 km with an average velocity of 2.52 m/s, due west, turns around, and hikes with an average velocity of 0.405 m/s, due east. How far east did she walk?
Solution:-
Given :-
Average velocity = 1.38m/s
Distance = 6.44km : converted km into m = 6440m
Average velocity due to west = 2.52m/s
Average velocity due to east = 0.405m/s
The average velocity is given by,
= Distance travelled / time taken
And its unit is meter/second (m/s)
Calculate time T1 = time she walked west,
= 6440 / 2.52
T1 = 2555 second
X = distance walked at east (m)
D = distance from starting point = 6440-x
T2 = time moving east = x / 0.405
Now distance (D)
D = (total tome )*(average velocity)
D = (T1 + T2 ) * ( 1.38)
6440-X = (2555 + X / 0.405) * 1.38
6440-X = ((1034.77 + X) / 0.405) * 1.38
6440-X = ( 1427.98 + 1.38 x) / 0.405
6440-x = 3525.87 + 3.4074 x
2914.13 = 3.4074x + 1x
2914.13 = 4.4074x
X = 2914.13 / 4.4074
X = 661.19m
So she walked 661.19m far east .