[Q1] Motion Type and Velocity Equations

The general kinematic equation for position under constant acceleration from an initial position of zero is $s(t) = v_0t + \frac{1}{2}at^2$. We can identify the motion type by comparing the given displacement equations to this standard form.

• Object A: The displacement is given by $s_A = 4t^2 + 2t$. Rearranging to match the standard form gives $s_A = 2t + 4t^2$. By comparison, we identify the initial velocity $v_{0A} = 2$ m/s and the acceleration term $\frac{1}{2}a_A = 4$ m/s², which yields a constant acceleration $a_A = 8$ m/s². Therefore, object A is undergoing uniformly accelerated motion. The corresponding velocity-time equation, $v(t) = v_0 + at$, is:

  $$v_A(t) = 2 + 8t$$

• Object B: The displacement is given by $s_B = 8t - 0.8t^2$. By comparison with the standard form, we identify the initial velocity $v_{0B} = 8$ m/s and the acceleration term $\frac{1}{2}a_B = -0.8$ m/s², which yields a constant acceleration $a_B = -1.6$ m/s². Therefore, object B is also undergoing uniformly accelerated motion. The corresponding velocity-time equation is:

  $$v_B(t) = 8 - 1.6t$$

[Q2] Time of Equal Velocities

To find the time when the velocities of the two objects are equal, we set their velocity equations equal to each other, $v_A(t) = v_B(t)$.

$$2 + 8t = 8 - 1.6t$$

We solve this equation for the time $t$:

$$8t + 1.6t = 8 - 2$$ $$9.6t = 6$$ $$t = \frac{6}{9.6}$$ $$t = 0.625 \, \text{s}$$

(Alternative Method using Calculus: Velocity is the time derivative of displacement, $v(t) = ds/dt$. Applying this, $v_A(t) = \frac{d}{dt}(4t^2 + 2t) = 8t + 2$ and $v_B(t) = \frac{d}{dt}(8t - 0.8t^2) = 8 - 1.6t$. These velocity equations are identical to those found above.)