[ r(t)=(2 t+1) i+(t-4) j ] * Displacement between t_(1)=2 s and t_(2)=4 s * Average velocity between t=2 s and t=4 * velocity at t=2 s * Acceleration at t=3 mathrm(~s)

To find the displacement between $t_1 = 2s$ and $t_2 = 4s$, we need to integrate the velocity function $P(t) = 2t + 1$ over the interval $[2, 4]$.<br /><br />The displacement is given by the integral of the velocity function over the given interval:<br /><br />$\text{Displacement} = \int_{2}^{4} (2t + 1) \, dt$<br /><br />Integrating the function, we get:<br /><br />$\text{Displacement} = \left[ t^2 + t \right]_{2}^{4}$<br /><br />Evaluating the integral, we have:<br /><br />$\text{Displacement} = (4^2 + 4) - (2^2 + 2) = 16 + 4 - 4 - 2 = 14$<br /><br />Therefore, the displacement between $t_1 = 2s$ and $t_2 = 4s$ is 14 units.<br /><br />To find the velocity at $t = 2s$, we simply substitute $t = 2$ into the velocity function $P(t) = 2t + 1$:<br /><br />$P(2) = 2(2) + 1 = 4 + 1 = 5$<br /><br />Therefore, the velocity at $t = 2s$ is 5 units.<br /><br />To find the acceleration at $t = 3s$, we need to take the derivative of the velocity function $P(t) = 2t + 1$:<br /><br />$P'(t) = 2$<br /><br />Since the derivative is a constant, the acceleration at any time $t$ is 2 units.<br /><br />Therefore, the acceleration at $t = 3s$ is 2 units.