A particle is located at 5i^ +6j^ + 7k^ at t=0.. After 5 seconds, it is located at 2i^+5j^ +5k^. The magnitude of displacement of  the particle in XY plane is: ???

Asked by nutankumari95074 | 11th Jun, 2022, 05:24: PM

Expert Answer:

Displacement = Final position - Initial position
 
Displacement = ( 2 begin mathsize 14px style i with hat on top end style + 5 begin mathsize 14px style j with hat on top end style+ 5 begin mathsize 14px style k with hat on top end style ) - ( 5 begin mathsize 14px style i with hat on top end style + 6 begin mathsize 14px style j with hat on top end style+ 7 begin mathsize 14px style k with hat on top end style ) = ( -3 begin mathsize 14px style i with hat on top end style - begin mathsize 14px style j with hat on top end style- 2 begin mathsize 14px style k with hat on top end style )
Magnitude of total displacement = begin mathsize 14px style square root of 3 squared plus 1 squared plus 2 squared end root space equals space square root of 14 space u n i t s end style
Magnitude of displacement in x-y plane = begin mathsize 14px style square root of 3 squared plus 1 squared end root space equals space square root of 10 space u n i t s end style

Answered by Thiyagarajan K | 11th Jun, 2022, 08:10: PM