A disoriented professor drives 5km east, 4km south, 2km west. Find the magnitude and direction of the resultant displacement.

Asked by lekhakarthikeyan | 8th Jun, 2017, 12:12: AM

Expert Answer:

begin mathsize 12px style The space man space travels space 5 space km space east semicolon space AB equals 5 space km
He space then space travels space 4 space km space south semicolon space BC equals 4 space km
Finally comma space he space travels space 2 space km space west comma space CD equals 2 space km
So comma space the space displacement space of space the space man space is space AD comma space AD equals straight x left parenthesis say right parenthesis
Draw space AE space and space DE space from space straight A space and space straight C comma space respectively.
From space the space geometry space of space the space figure comma
AE equals 4 space km space and space DE equals 5 minus 2 equals 3 space km
From space Pythagoras apostrophe space Theorem comma
straight x squared equals 3 squared plus 4 squared equals 25
therefore straight x equals 5 space km
Hence comma space the space displacement space of space the space man space is space 5 space km

Let space us space say space that space the space angle space EAD space be space straight beta
therefore tanβ equals ED over AE equals 3 over 4 equals 0.75
therefore straight beta equals tan to the power of negative 1 end exponent open parentheses 0.75 close parentheses equals 36.87 degree
Let space the space direction space of space displacement space make space an space angle space straight theta space with space the space horizontal space left parenthesis AB right parenthesis.
therefore straight theta equals 90 minus 36.87 equals 53.13 degree
Hence comma space the space displacement space is space 5 space km space and space 53.13 degree space south space of space east. end style

Answered by Romal Bhansali | 8th Jun, 2017, 09:20: AM

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