Question

26. An aeroplane moves 400 m towards north turns
300 m towards west and then 1200 m vertically
upward. Then its displacement from initial
position :
(1) 1900 m
(3) 1500 m
(2) 1600 m
(4) 1300 m​

Answer 1

Answer:

it's displacement

$= \sqrt{(\sqrt{400 {}^{2} + 300 {}^{2} } ) {}^{2} + 1200 {}^{2} } m \\ = \sqrt{1690000} \\ = 1300m$

Answer 2

Answer:

1300m

Explanation:

Refer figure above

• Figure 1: First the plane moves from A to B which is North 400m
• Then it moves West which is to point C that is 300m. Using Pythagoras theorem displacement between A and C becomes : square root (400^2 + 300^2) = 500
• Figure 2: Now the plane moves vertically upward which doesn't mean it flies north rather it goes into a third direction which assume is like coming towards you (out of the paper) which is indicated by 'X'
• Now it essentially means you get another triangle where it moves from C to D (towards you)
• We get triangle ACD and final displacement is the length AD which can be calculated using Pythagoras theorem and comes out to be 1300m