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A monkey climbs up a slippery pole for 3 and subsequently slips for 3. Its velocity at time t is given by v (t) = 2t (3 - t), 0 ltt lt 3 and v(t) = -(t - 3) (6-t) for 3 lt tlt 6 s in m/s. It repeats this cycle till it reaches the height of 20 m. (a) At what time is its velocity maximum ?(b) At what time is its average velocity maximum ? (c) At what time is its acceleration maximum in magnitude ? (d) How many cycles (counting fractions) are required to reach the top ? |
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Answer» Solution :For maximum velocity, `(dv)/(dt) = 0` Given velocity, `v = 2t (3-t) = 6t - 2t ^(2)` (a) From maximum velocity, `(dv)/(dt) =0` `THEREFORE (d)/(dt) (6t - 2t ^(2)) =0` `therefore 6- 4t =0` `therefore t = (6)/(4) = 3/2 s = 1.5 s` (b) From equ. (i) `v = 6t - 2t ^(2)` `therefore (dx)/(dt) = 6t - 2t ^(2)` `therefore (dx)/(dt) = 6t - 2t ^(2)` `therefore dx = (6t - 2t ^(2)) dt` `therefore` Distance travelled in time interval 0 to 3 s. `x _(1) = int _(0) ^(3) ( 6t - 2t ^(2)) dt` `= [ ( 6t ^(2))/( 2 ) - ( 2t ^(3))/( 3) ] _(0) ^(3) = [ 3t ^(2) - (2)/(3) t ^(3) ] _(0) ^(3)` `= 3 XX 9 - 2/3 xx 3 xx 3 xx 3=27-18 =9m` Average velocity `= ("Displacement")/("Time")=9/3=3m//s` Given, `x = 6t - 2t ^(2)` `3 = 6t - 2t ^(2)` `therefore 2t ^(2) - 6t - 3=0` `therefore t = (6pm sqrt (6 ^(2) - 4 xx 2 xx 3 ))/(2 xx 2 ) = ( 6pm sqrt ( 36 - 24))/( 4)` `= (6pm sqrt12)/(4) = (3pm 2 sqrt3)/(2)` CONSIDERING positive sign only. `therefore t = (3 + 2 sqrt3)/( 2) = (3 + 2 xx 1.732)/(2) = 9/4s` (c) In a periodic motion when velocity is zero acceleration will be maximum PUTTING v =0 in equ. (i), `0= 6t - 2t ^(2)` ` therefore 0= t ( 6- 2t)` ` = t xx 2 (3-t) =0` `therefore t = 0 or t =3 s` (d) Distance covered in 0 to `3s = 9m` Distance covered in 3 to `6s int _(3) ^(5) (18-9t +t ^(2)) dt` `x _(2) = (18 t - (9t^(2))/( 2) + (t ^(3))/( 2)) _(3) ^(6)` `= 18 xx 6- 9/2xx 6^(2) + (6 ^(3))/(3)` `- (18 xx 3 - (9 xx 3 ^(2))/(2) + ( 3 ^(3))/( 3))` `= 108 -9 xx 18 + (6 ^(3))/( 3) ` `- 18 xx 3 + 9/2 xx 9- (27)/(3) = 108 - 18 xx 9+ (216)/(3) ` `- 54 + 4.5 xx 9-9 =- 4.5 m` `therefore` Total distance travelled in oen cycle, `= x _(1) + x _(2) = 9 -4.5 = 4.5 m` Number of cycles covered, `= ("total distance ")/("distance covered in each cycle")= (20)/(4.5) ~~4.44 ~~5` |
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