The following figure is the speed-time graph for a rocket from the moment when the fuel starts to burn,i.e.,at time $t=0$.
$(a)$ State the acceleration of the rocket at $t=0$.
$(b)$ State what happens to the acceleration of the rocket between $t=5 \, s$ and $t=60 \, s$.
$(c)$ Calculate the acceleration of the rocket at $t=80 \, s$. Give a reason for your answer.
$(d)$ The total mass of the rocket at $t=80 \, s$ is $1.6 \times 10^{6} \, kg$. Calculate the resultant force on the rocket at this time. Give a reason for your answer.

(N/A) At $t=0$,the speed is $0$ and the slope of the graph is $0$. Therefore,the initial acceleration is $0 \, m/s^2$.
$(b)$ Between $t=5 \, s$ and $t=60 \, s$,the slope of the speed-time graph is increasing. Since the slope of a speed-time graph represents acceleration,the acceleration of the rocket is increasing during this interval.
$(c)$ At $t=80 \, s$,the graph is a straight line. The slope of this line is constant. Calculating the slope between $t=60 \, s$ $(v \approx 480 \, m/s)$ and $t=100 \, s$ $(v = 1400 \, m/s)$: $a = \frac{1400 - 480}{100 - 60} = \frac{920}{40} = 23 \, m/s^2$. The acceleration is constant because the graph is linear in this region.
$(d)$ Using Newton's second law,$F = ma$. Given $m = 1.6 \times 10^{6} \, kg$ and $a = 23 \, m/s^2$,the resultant force $F = 1.6 \times 10^{6} \times 23 = 3.68 \times 10^{7} \, N$. The force is non-zero because the rocket is accelerating.