Question

A particle is moving eastwards with a velocity of 5 m/s in 10 seconds the velocity! northwards. The average acceleration in this time is 45 mis in 10 seconds the velocity changes to 5 ods [AIEEE-2005] (A) m/s towards north-east (B) -m/stowards north. (C) zero L (D) Tam/s² towards north-west

JEE/Engineering Exams

Physics

Solution

175

4.0 (1 ratings)

Intial velocity ( =v 1=5 mathrm{m} / ) seast wards [ wedge |^{wedge} ] then after time t=10sec Il towards north ( 5 mathrm{m} / mathrm{s} ) ( v 2=5 m / s ) north wards [ Rightarrow Rightarrow Rightarrow Rightarrow Rightarrow ] towards east ( 5 mathrm{m} / mathrm{s} ) Vresultant ( =sqrt{v} 1^{2}+v 2^{2} ) [ begin{array}{l} =sqrt{2} 5+25 =5 sqrt{2} end{array} ] acceleration a=velocity/time [ begin{array}{l} =5 sqrt{2} / 10 =sqrt{2} / 2 times sqrt{2} / sqrt{2} end{array} ] ( =1 / sqrt{2} ) towards ( n-w )