InterviewSolution
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InterviewSolution
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A particle is moving eastwards with a velocity of ` 5 m//s`. In `10 s` the velocity changes to `5 m//s` nothwards. The average acceleration in this time isA. ` 1 /(sqrt 2) ms^(-2)` , 45 West of NothB. (b) ` 1/2 ms^(-2), 60^@ west of NorthC. ` 2 ms^(-2) 60^@` East of soutD. `1/(sqrt 2) ms^(-2) `, 30^@` West of south |
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Answer» Correct Answer - A Here, initial velocity ` v_1 = 5 ms^(-1)` along East ` = 5 hat i` ` vec _1 = 5 ms^(-1)` along North `= 5 hat j` Change in velocity, ` Deltav= vec v_2 = vec v_1 = 5 hat j - 5 hat i=- 5 hat I + 5 hat j` Magnitude of ` Delta v= sqrt ((-5) ^2 + 5^2 ) = 5 sqrt 2 m//s` Acceleration, ` |vec a| = (Delta v0/t = (5 sqrt 2)/(10) = 1/(sqrt 2) ms^(-2)` ` tan theta = (BD)/(OB) = 5/ 5 = 1 tan 45^@` ` theta 45^2` West of Noth`. |
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