InterviewSolution
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InterviewSolution
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Two cars (A) and (B) travel in straight line. The distance of (A) from the starting point is given as a function of them be ` a_A (t) = pt + qt^2 `, with ` p = 2 . 60 ms^(-1)` and ` q= 1.20 ms^(-2)`. The distance of (B) from the starting point is ` x_B (t) = rt^2 - st^3 ` are ` r= 2. 80 ms^(-2)` and ` s = 0.20 ms^(-3)`. Answer the following questions, At what time (s) are the cars at the same point ?A. ` 2. 60`B. ` 2. 27 s`C. ` 5. 7 3 s`D. ` both ` 2. 27 `and ` 5 . 73 s` |
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Answer» Correct Answer - D ` x_A=A_b` :. Pt+qt^2 =rt^2 -st^3` One solution is ` t= 0`, which means both the cars start from the same point. To find the other solution , divede it by ` 9t), we have ` p + qt =rt -st^2` or ` st^2 + (q-2) t + p=0` or ` t= 1/(2 s) [ q- r) =- sqrt((q -r)^2 -4sp)]` ` =1/(2 xx 0.20) [1 . 20 - 2 . 89 )` ` +- sqrt ((1.20 - 2 .80)^2 - 4 xx 0.24 xx 2.60)]` `= 4. 00 +- 73 = 2.27 s ` and ` 5. 73 s`. Hence ` x_A= x_B` for ` t= 0, t= 2.27 s` and ` t= 5. 73 s`. |
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