InterviewSolution
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InterviewSolution
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A force F is applied to the initially, stationary cart. The variation of force with time is shown in the figure. The speed of cart at t = 5 s is A. 10 m/sB. 8.33 m/sC. 2 m/sD. zero |
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Answer» Correct Answer - B The equation of parabola is `x^(2)=4ay` Here, `t^(2)=4aF` When t = 5 s, F = 50 N (see graph) `therefore" "5^(2)=4axx50` `therefore " "a=(25)/(200)=(1)/(8)` `therefore t^(2) = 4 xx(1)/(8)F=(F)/(2)implies F=2t^(2)` Acceleration `=(F)/(m)=(2t^(2))/(m)=(2t^(2))/(10)=(t^(2))/(5)` `therefore(dv)/(dt)=(t^(2))/(5)" or "int_(0)^(v)dv=int_(0)^(5)(t^(2))/(5)dt` `therefore v = (1)/(5)[(t^(3))/(3)]_(0)^(5)=(125)/(15)=8.33 m//s` |
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