A monkey climbs up a slippery pole for 3 seconds and subsequently slips for 3 seconds. Its velocity at time t is given by v(t) = 2t (3-t); 0< t < 3 and v (t)=–(t–3)(6–t) for 3 < t < 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?

11 Views|Posted 7 months ago
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7 months ago

This is a long answer type question as classified in NCERT Exemplar

(a)  for maximum velocity dv/dt=0

d/dt (6t-2t2)=0

6-4t=0 t= 6/4=1.5s

(b) v=6t-2t2

ds/dt=6t-2t2

ds=6t-2t2dt

distance in 3s, S= 0 3 6 t - 2 t 2 d t = [ 3 t 2 - 2 3 t 3 ] 30

s= 27-18=9m

average velocity = distance /time =9/3 = 3m/s

x= 6t-2t2

3=6t-2t2

After solving we get t= 9/4s approx

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According to question, we can write

 10 = 1 2 a t 2 . . . . . . . . . . . . . . . ( 1 ) a n d                        

1 0 + x = 1 2 a ( 2 t ) 2 . . . . . . . . . . . . . . . . ( 2 )  

X = 1 2 A [ ( 2 t ) 2 t 2 ] = 3 ( 1 2 a t 2 ) = 3 0 m  

 

Average speed = 4 v 2 3 v

= 4 v 3

(d) Initial velocity  = - v j ˆ

Final velocity = v i ˆ

Change in velocity  = v i ˆ - ( - v j ˆ )

= v ( i ˆ + j ˆ ) Momentum gain is along i ˆ + j ˆ

 Force experienced is along i ˆ + j ˆ

  Force experienced is in North-East direction.

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Physics NCERT Exemplar Solutions Class 11th Chapter Three 2025

Physics NCERT Exemplar Solutions Class 11th Chapter Three 2025

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