On the dog's return trip (between <em>t</em> = 10 and <em>t</em> = 12.5 seconds), the slope of the position function is steeper than during the first 5 seconds, which means the dog ran home faster. The only option that captures this is D.
You can check to make sure that the dog indeed runs twice as fast on the return trip. The slope of the position function during the first 5 seconds is
(change in position) / (change in time) = (5 - 0) / (5 - 0) = 5/5 = 1
while during the return trip, it is
(0 - 5) / (12.5 - 10) = -5/2.5 = -2
Ignoring the sign (which only indicates the direction in which the dog was running), we see that the dog's speed on the return trip was indeed twice as high as during the first 5 s.
