$A$ certain force exerted for $1.2 \, s$ raises the speed of an object from $1.8 \, m s^{-1}$ to $4.2 \, m s^{-1}$. Later, the same force is applied for $2 \, s$. How much does the speed change in $2 \, s$?

(D) Given:
Time interval, $t_1 = 1.2 \, s$
Initial speed, $u_1 = 1.8 \, m s^{-1}$
Final speed, $v_1 = 4.2 \, m s^{-1}$
Using the first equation of motion, $v = u + at$, we calculate the acceleration $(a)$:
$a = \frac{v_1 - u_1}{t_1} = \frac{4.2 - 1.8}{1.2} = \frac{2.4}{1.2} = 2 \, m s^{-2}$
Since the same force is applied to the same object, the acceleration remains constant at $a = 2 \, m s^{-2}$.
For the second interval, $t_2 = 2 \, s$:
The change in speed $(\Delta v)$ is given by $\Delta v = a \times t_2$.
$\Delta v = 2 \, m s^{-2} \times 2 \, s = 4 \, m s^{-1}$.
Thus, the speed changes by $4 \, m s^{-1}$.