Question

5. A boy A is standing 3 m west and 4 m north to a boy B. Assume the east and the north towards the
positive x and y-axis of coordinate system. The boy A starts moving along a vector a = 1.5i +2j with a
constant speed of 2 m/s for 5 s and stops. Find its new position vector with respect to the boy B. -Ism
8. A force P of 235 N acts at an angle one with the horizontal on hick stinn
990 balone​

Answer 1

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

Regardsone year 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

= 15m