a man goes 10 meter towards north then 20 meter towards east then the displace ment is ?
please send answer with steps
Adityaadidangi:
√(20²+10²)
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hey!!!
we know the shortest distance between the initial point and final point is called displacement.
according to the question, first the man goes 10m towards north. let the starting point be a.
and then 20m towards east. final point be b.
and displacement be c.
by Pythagoras thereom :-
>> a² + b² = c²
>> 10² + 20² = c²
>> 100 + 400 = c²
>> 500 = c²
>> c = √500m = 10√5m
hence, displacement is 10√5ms^-1
cheers!!!
we know the shortest distance between the initial point and final point is called displacement.
according to the question, first the man goes 10m towards north. let the starting point be a.
and then 20m towards east. final point be b.
and displacement be c.
by Pythagoras thereom :-
>> a² + b² = c²
>> 10² + 20² = c²
>> 100 + 400 = c²
>> 500 = c²
>> c = √500m = 10√5m
hence, displacement is 10√5ms^-1
cheers!!!
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answer photo pe hai
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