InterviewSolution
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InterviewSolution
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A particle starts from the origin at ` t=0` with a velocity of ` 10.0 hat j m//s` and moves in the X-y` plane with a constant accleration of ` ( 8.0 hat I + 2. 0 hat j ) ms^(-2) . (a) At wht time is the s-coordinate of the particle ` 16 m` ? What is the y-coordinate of the particle at that time ? ( b) What is the speed of the particle at that time ? |
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Answer» Here, ` vec u= 10.0 hat j ms^(-1) at t=0`. ` vec a = (d vec v)/(dt) = ( 8. 0 hat I + 2.0 hat j) ms^(-2)` So ` d vec v =( 8. 0 hat I + 2.0 hat j) t ` or ` vec v = vec u + 8.0 t hat I + 2.0 t hat j` As` vec v= (d vec r)/(dt) ` or ` d vec r = vec v dt` so, ` d vec r = (vec u + 8.0 t hat + 2.0 t hat j ) dt` Integrating it within the conditions of motion i.e. as time changes from ` 0` to `t`, displacement changes from ` 0` to ` r`, we have : ` vec r = vec u t + 1/2 xx 8.0 t ^2 hat i + 1/2 xx 2.0 t^2 hat j ` or ` z hat i + y hat j = t + 4. 0 t ^2 hat i + t^2 hat j = 4.0 t^2 hat i + (10 t + t^2) hat j` Here, qwe have, ` = 4.0 t^2` and ` y=10 t+ t^2` :. ` t= ( x//4) ^(1//2)` (b) Velocity of the particle at time (t) is ` vec v= 10 hat j 8.0 t +2.0 t hat j` When ` t= 2 s`, then, ` vec v = 10 hat j + 8.0 xx 2 hat i + 2.0 xx 2 hat j = 16 hat i + 14 hat j` Speed` = | vec v| = sqrt ( 16 ^@ + 14^@ ) = 21. 26 ms^(-10`. |
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