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A monkey climbs up a slippery pole for ...

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 ?

A

1.5 s

B

3 sec

C

2s

D

5s

Text Solution

Verified by Experts

The correct Answer is:
A
For maximum velocity `(dv(t))/(dt)=0`
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Knowledge Check

  • The velocity of the particle at any time t is given by vu = 2t(3 - t) m s^(-1) . At what time is its velocity maximum?

    A
    2 s
    B
    3 s
    C
    2/3 s
    D
    3/2 s

  • Displacement x at time t is x=t-t^(3) . If v is the velocity and a the acceleration at time t, then :v =

    A
    `(1)/(2)at^(2)`
    B
    `1+(1)/(2) at`
    C
    `ax`
    D
    `a+(1)/(2)t`

  • Velocity of a particle is given as v = (2t^(2) - 3)m//s . The acceleration of particle at t = 3s will be :

    A
    `18 ms^(-2)`
    B
    `12 ms^(-2)`
    C
    `15 ms^(-2)`
    D
    zero
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