Answer
The more stops there are, the time taken is greater and the train's average speed is slower.
Work Step by Step
$95\frac{km}{hr}\times \frac{1 hr}{3600s}\times \frac{1000m}{1km}=26 \frac{m}{s}$
a)Acceleration: 1.1 $\frac{m}{s^2}$
$t=\frac{26\frac{m}{s}-0\frac{m}{s}}{1.1\frac{m}{s^2}}=24s$
$\Delta x=\frac{26\frac{m}{s}+0\frac{m}{s}}{2}\times 24s=312m$
Deceleration: -2.0$\frac{m}{s^2}$
$t=\frac{0\frac{m}{s}-26\frac{m}{s}}{-2.0\frac{m}{s^2}}=13s$
$\Delta x=\frac{0\frac{m}{s}+26\frac{m}{s}}{2}\times 13s=169m$
$3000m-312m-169m=2519m$
$t=\frac{2519m}{26\frac{m}{s}}=97s$
In part a, there are 5 acceleration, 5 constant speed, 5 deceleration phases, and 4 22 second stopping periods. The total time needed to complete the track is
$5\times (24s+97s+13s)+4\times22s=758s$
Average speed: $\frac{15000m}{758s}=19.8\frac{m}{s}$
b) Here, the only different time interval is the constant speed interval.
$5000m-312m-169m=4519m$
$t=\frac{4519m}{26\frac{m}{s}}=170s$
In part b, there are 3 acceleration, 3 constant speed, 3 deceleration phases, and 2 22 second stopping periods. The total time needed to complete the track is
$3\times (24s+170s+13s)+2\times22s=665s$
Average speed: $\frac{15000m}{665s}=22.6\frac{m}{s}$
We have shown that the more stops there are, the greater time it takes to complete and the slower the train's average speed.
