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Velocity (in m/s) of a particle moving a...

Velocity (in m/s) of a particle moving along x-axis varies with time as, `v= (10+ 5t -t^2)` At time `t=0, x=0.` Find
(a) acceleration of particle at `t = 2 s` and
(b) x-coordinate of particle at `t=3s`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
(a) `a=(dv)/(dt)=5-2t`
At `t=2s`
`a=1 ms^2`
(b) `x=int_0^3 v dt =int_0^3(10+5t-t^2)dt`
`=[10t+2.5t^2-t^3/3]_0^3`
`=(10xx3)+(2.5)(3)^2-(3)^3/3`
`=43.5 m`
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Knowledge Check

  • The position (x) of a particle moving along x - axis veries with time (t) as shown in figure. The average acceleration of particle in time interval t = 0 to t = 8 s is

    A
    `3m//s^(2)`
    B
    `-5m//s^(2)`
    C
    `-4m//s^(2)`
    D
    `2.5m//s^(2)`

  • v_(x) is the velocity of a particle moving along the x-axis as shown in the figure. If x=2.0m at t=1.0s, what is the position of the particle at t=6.0s ?

    A
    `-2.0` m
    B
    `+2.0` m
    C
    `+1.0` m
    D
    `-1.0` m

  • the angular velocity omega of a particle varies with time t as omega = 5t^2 + 25 rad/s . the angular acceleration of the particle at t=1 s is

    A
    `10 rad/s^2`
    B
    `5rad/s^2`
    C
    Zero
    D
    `3 rad/s^2 `