InterviewSolution
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InterviewSolution
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A monkey climbs up a slippery pole for ` 3 seconds` and subsequently slips for `3 seconda`. Its velocity at time (t) is given by ` v (t) =2 t (3-t) , 0 lt tgt 3 s and v (t) =- (t-30 (6-t) for 3 lt tlt 6 s in m//s`. It repeats thei cycle till it reaches the height of ` 20 s.` (a) AT wht time is its v elocity maximum ? (b) At what time is its average velocity maximum ? (c ) At what time is its accelration maximum in magnitude ? (d) How many cycles (counting fractions ) are required to reach the top ? |
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Answer» It this problem to calculate maximum velocity we will use `(dv)/(dt) =0`, then the time corresponding to maximum velocity will be obntained. Given velocity `v(t) = 2t (3-t) = 6t - 2t^(2)"…….."(i)` (a) For maximum velocity `(dv(t))/(dt) = 0` `rArr d/(dt) (6t - 2t^(2)) = 0` `rArr 6 - 4t =0` `rArr t = (6)/(4) = (3)/(2)s = 15s` (b) From Eq. (i) `v = 6t - 2t^(2)` `rArr (ds)/(dt) = 6t - 2t^(2)` `rArr ds = (6t - 2t^(2))dt` where,s is displacement `:.` Distane travelled in time interval 0 to 3s. `s = int_(0)^(3) (6t - 2t^(2)) dt` `= [(6t^(2))/(2) - (2t^(3))/(3)]_(0)^(3) = [3t^(3) - (2)/(3)t^(3)]_(0)^(3)` `= 3 xx 9 - (2)/(3) xx 3 xx 3 xx 3` `= 27 - 18 = 9 m` Average velocity ` = ("Distance travelled")/("Time")` `= 9/3 = 3 m//s` Given, `x = 6t - 2t^(2)` `rArr 3 = 6t - 2t^(2)` `rArr 2t^(2) - 6t - 3 = 0` `rArr t = (6 +- sqrt(6^(2) - 4 xx 2 xx 3))/(2 xx 2) = (6 +- sqrt(36-24))/(4)` `= (6+- sqrt(12))/(4) = (3+- 2sqrt(3))/(2)` (c) In a periodic motion when velocinty is zero accleration will be maximum putting `v = 0` in Eq. (i) `0 = 6t - 2t^(2)` `rArr 0 = t (6-2t)` `= t xx 2 (3-t) = 0` `rArr t = 0` or `3s` (d) Distance covered in `0` to `3s = 9m` Distance covererd in 3 to 6s = `int_(3)^(6) (18 - 9t +t^(2)) dt` `= (18t - (9t^(2))/(2) + (t^(3))/(3))_(3)^(6)` `= 18 xx 6 - (9)/(2) xx 6^(2) + (6^(3))/(3) - (18 xx 3 - (9 xx 3^(2))/(2) + (3^(3))/(3))` `= 108-9xx18+(6^(3))/(3) - 18 xx 3 + (9)/(2) xx 9 - (27)/(3)` `= 108-18 xx 9 + (216)/(3) - 54+45xx9-9 = - 45 m` `:.` Total distance travelled in one cycle `= s_(1) + s_(2) = 9 - 45 = 45m` Number of cycles covered in total distance to be covered `= (20)/(4.5) = 4.44 ~~ 5` |
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