Consider a particle moving along x-axis. Its distance from origin O is described by the co-ordinate x which varies with time. At a time `t_(1)`, the particle is at point P, where its co-ordinate is `x_(1)` and at time `t_(2)`, the particle is at point Q, where its co-ordinate is `x_(2)`. The displacement during the time interval from `t_(1)` to `t_(2)` is the vector from P to Q, the x-component of this vector is `(x_(2) - x_(1))` and all other components are zero. It is convenient to represent the quantity `x_(2) - x_(1)` the change in x by means of a notation `Delta`, thus `Delta x = x_(2) - x_(1)` and `Delta t = t_(2) - t_(1)`. The average velocity `bar(V) = (x_(2) - x_(1))/(t_(2) - t_(1)) = (Delta x)/(Delta t)` A particle moves according to the equation `x = t^(2) + 3t + 4`. The average velocity in the first 5 s is:A. 8 m/sB. 7.6 m/sC. 6.4 m/sD. 5.8 m/s

Consider a particle moving along x-axis. Its distance from origin O is

1.	Consider a particle moving along x-axis. Its distance from origin O is described by the co-ordinate x which varies with time. At a time `t_(1)`, the particle is at point P, where its co-ordinate is `x_(1)` and at time `t_(2)`, the particle is at point Q, where its co-ordinate is `x_(2)`. The displacement during the time interval from `t_(1)` to `t_(2)` is the vector from P to Q, the x-component of this vector is `(x_(2) - x_(1))` and all other components are zero. It is convenient to represent the quantity `x_(2) - x_(1)` the change in x by means of a notation `Delta`, thus `Delta x = x_(2) - x_(1)` and `Delta t = t_(2) - t_(1)`. The average velocity `bar(V) = (x_(2) - x_(1))/(t_(2) - t_(1)) = (Delta x)/(Delta t)` A particle moves according to the equation `x = t^(2) + 3t + 4`. The average velocity in the first 5 s is:A. 8 m/sB. 7.6 m/sC. 6.4 m/sD. 5.8 m/s
Answer» Correct Answer - A `x = t^(2) + 3t + 4` `x_(1) = 4` `x_(2) = 44` `v_(av) = ((x_(2) - x_(1))/(Delta T))`

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