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Velocity of a particle varies as v=2t^3-...

Velocity of a particle varies as `v=2t^3-3t^2` in `(km)/(hr)` If `t=0` is taken at 12:00 noon
Q. Find the time between 12:00 noon and 1:00 pm which speed is maximum

A

12:00 noon

B

0.54166666666667

C

0.45833333333333

D

0.58333333333333

Text Solution

Verified by Experts

The correct Answer is:
A
`v=2t^3-3t^2`
`(dv)/(dt)=a=6t(t-1)`
`(dv)/(dt)=0`
`t=0`, 1sec
`(d^2v)/(dt^2)=12t-6implies((d^2v)/(dt^2))_(t=0)=-6`
`((d^2v)/(dt^2))_(t=1)=6`
At `t=0`, The sigh of double derivative is negative hence velocity will be maximum at 12:00 noon. At `t=1`, the sign of double derivative is positive hence the speed will be minimum at 1:00
P.M.
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Knowledge Check

  • Velocity of a particle varies as v=2t^3-3t^2 in (km)/(hr) If t=0 is taken at 12:00 noon Q. What is the velocity of the particle at 12:00 noon?

    A
    `0.5 (km)/(hr)`
    B
    `zero
    C
    1km `(km)/(hr)`
    D
    2 `(km)/(hr)`

  • Velocity of a particle varies as v=2t^3-3t^2 in (km)/(hr) If t=0 is taken at 12:00 noon Q. The time at which speed of the particle is minimum.

    A
    12:00 noon
    B
    0.54166666666667
    C
    0.45833333333333
    D
    0.58333333333333

  • Velocity of a particle varies as v=2t^3-3t^2 in (km)/(hr) If t=0 is taken at 12:00 noon Q. Find the expression for the acceleration of the particle.

    A
    `3t^2+3t`
    B
    `6t(t-1)`
    C
    `6t^2+3t`
    D
    none

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