KVPY 2016 SA Question 8

Classical Mechanics Level 3

A person walks 2 5 ∘ 25^\circ north of east for 3.18 km. How far should she have to walk due north and then east to reach the same position?

Try my set KVPY 2016 SA Questions .
Towards 1.24 km North and 2.78 km East Towards 1.34 km North and 2.88 km East Towards 2.88 km North and then 1.34 km East Towards 1.64 km North and 2.51 km East

Md Zuhair by Md Zuhair , 20, India

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1 solution

Sharon Walter
Nov 4, 2017
  • The person walks 25 degree NORTH OF EAST. In order to reach the same position by first moving EAST and then NORTH, it will form a right angled triangle with the original distance as the hypotenuse - 3.18 km, right angled at the point where the person turns to north.

therefore, 3.18 is the hypotenuse of the right angles triangle hence, by pythagorus theorem, if we calculate the sums of the the squares of each value given in the option, the closest value to the square of the hypotenuse shall the correct option.

i.e. option c and d since, the square of 3.18=10.11 and the the sum of the square of 2.88 and 1.34 = 10.09

Now, we have to think practically here, we have considered a right angled triangle, so if the direction towards the east is smaller than north, then a right angled triangle will not be formed, since the length towards north will go diagonal in order to meet the final position of the person

therefore we can say that, direction towards the east > directs towards the north

hence option D is Correct

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