Question

In: Physics

In reaching her destination, a backpacker walks with an average velocity of 1.38 m/s, due west....

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?

Expert Solution

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 .

genius_generous answered 1 year ago

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