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boy a is standing at a point 3 m west and 4 m south to boy b. assume the east and the north towards the positive x and y axis of coordinate system. boy a starts moving along a vector a =1.5 i +2 j with a constant speed of 2m/s for 5 seconds and stops. now how far is the boy a from the boy b

Gaurav Adhikari , 5 Years ago
Grade
anser 3 Answers
Saurabh Koranglekar

Last Activity: 5 Years ago

Dear student

Motion is in angle tan inverse 4/3 = 53 degree

Therefore speed in x direction 2 cos (53)
And speed in y direction 2 sin ( 53)

Therefore distance in 5 seconds = 6 i + 8 j

Distance moved from -3i -4j is 3i + 4j

And therefore distance moved from origin or boy b is 3i + 4j


Regards

Khimraj

Last Activity: 5 Years ago

boy A is standing at a point 3 m west and 4 m south to boy B. Assume the east and the north towards the positive x and y axis of coordinate system. boy a starts moving along a vector a =1.5 i +2 j with a constant speed of 2m/s for 5 seconds and stops. now how far is the boy A from the boy B ?
let say boy B is standing at origin (0,0)
then boy A is standing (-3 , -4)
boy A move 1.5i + 2 j 
distance moved = 2 × 5 = 10m
1.5i + 2j = root (1.5^2 + 2^2) = 2.5
distance moved toward i = east =(1.5)/(2.5) × 10 = 6 m
distance moved toward j = north = (2/2.5) × 10 = 8m
BoyA = ( 6, 8)
Boy B = (-3 , -4)
distance between boy A& B = root ( (6-(-3))^2 + (8 -(-4))^2)
= root ( 9^2 + 12^2)
= root ( 81 + 144)
= root 225
 = 15 m 

Shreekant Sharma

Last Activity: 2 Years ago

Motion is in angle tan inverse 4/3 = 53 degree
 
Therefore speed in x direction 2 cos (53)
And speed in y direction 2 sin ( 53)
 
Therefore distance in 5 seconds = 6 i + 8 j
 
Distance moved from -3i -4j is 3i + 4j
 
And therefore distance moved from origin or boy b is 3i + 4j
Therefore distance travelled by A from b is under root of 4²+3² which is 5 
Hence 5 is the answer 
 

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