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:

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:
Answer» 8 m/s 7.6 m/s 6.4 m/s 5.8 m/s Solution :`x = t^(2) + 3T + 4` `x_(1) = 4` `x_(2) = 44` `v_(av) = ((x_(2) - x_(1))/(DELTA T))`