InterviewSolution
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InterviewSolution
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Let `r_(1)(t)=3t hat(i)+4t^(2)hat(j)` and `r_(2)(t)=4t^(2) hat(i)+3t^(2)hat(j)` represent the positions of particles 1 and 2, respectiely, as function of time t, `r_(1)(t)` and `r_(2)(t)` are in metre and t in second. The relative speed of the two particle at the instant t = 1s, will beA. 1 m/sB. `3sqrt(2) m//s`C. `5sqrt(2) m//s`D. `7sqrt(2)m//s` |
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Answer» Correct Answer - C Given, `r_(1)=3t hat(i)+4t^(2)hat(j)` `therefore (dr_(1))/(dt)=3hat(i)+8t hat(j)` At t = 1 s `upsilon_(1)=(dr_(1))(dt)=3 hat(i)+8hat(j)` Again, `r_(2)(t)=4t^(2)hat(i)+3hat(t)hat(j)` `(dr_(2))/(dt)=8hat(t)hat(j)+3hat(j)` At t = 1 s `upsilon_(2)=(dr_(2))/(dt)=8hat(i)+3hat(j)` Relative velocity `=upsilon_(1)-upsilon_(2)=-5hat(i)+5hat(j)` `=sqrt((5)^(2)+(5)^(2))=5sqrt(2)ms^(-1)` |
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