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A particle moves in the x-y plane accord...

A particle moves in the x-y plane according to the law x=at, y=at `(1-alpha t)` where a and `alpha` are positive constants and t is time. Find the velocity and acceleration vector. The moment `t_(0)` at which the velocity vector forms angle of `90^(@)` with acceleration vector.

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AI Generated Solution

To solve the problem, we will follow these steps: ### Step 1: Write the position vector The position of the particle in the x-y plane is given by: \[ x = at \] \[ y = at(1 - \alpha t) \] Thus, the position vector \( \mathbf{r} \) can be expressed as: \[ \mathbf{r} = x \hat{i} + y \hat{j} = at \hat{i} + at(1 - \alpha t) \hat{j} \] ...
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Knowledge Check

  • A particle moves in x-y plane according to the equations x= 4t^2+ 5t+ 16 and y=5t where x, y are in metre and t is in second. The acceleration of the particle is

    A
    `8 m s^(-2)`
    B
    `12 m s^(-2)`
    C
    `14 m s^(-2)`
    D
    `16 m s^(-2)`
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