The diagram shows the speed-time graph of a train for the first 62.5 seconds after entering a tunnel.

1. Find the constant rate of change of speed on the first straight-line segment.

From $t=0$ to $t=25$, the speed decreases from $100\text{ m/s}$ to $20\text{ m/s}$.

$\text{Gradient}=\frac{20-100}{25-0}=\frac{-80}{25}=-3.2\text{ m/s}^2$

2. Find the speed after 20 seconds in m/s.

At $t=20$,

$\text{speed}=100-3.2(20)=36\text{ m/s}$

3. Convert the speed from m/s to km/h.

Since $1\text{ m/s}=3.6\text{ km/h}$,

$36\times 3.6=129.6\text{ km/h}$

1. Identify distance as the area under the speed-time graph from $t=0$ to $t=25$.

Distance travelled $=$ area under the speed-time graph.

From $0$ to $25$ seconds, the graph forms a trapezium with parallel sides $100$ and $20$, and width $25$.

2. Calculate the area of the trapezium.

$\text{Distance}=\frac{1}{2}(100+20)(25)$
$=60\times 25$
$=1500$

1. Find the deceleration during the first 25 seconds.

From $t=0$ to $t=25$, the speed decreases from $100$ m/s to $20$ m/s.

$\text{deceleration}=\frac{100-20}{25}=\frac{80}{25}=3.2\text{ m/s}^2$

2. Use the given relationship to find the acceleration after 25 seconds and calculate $k$.

The deceleration is twice the acceleration after $25$ seconds.

$3.2=2\times \text{acceleration}$

$\text{acceleration}=1.6\text{ m/s}^2$

From $t=25$ to $t=62.5$, the time taken is

$62.5-25=37.5\text{ s}$

So

$\frac{k-20}{37.5}=1.6$

$k-20=1.6\times 37.5=60$

$k=80$