What is the answer for question 2. b)? http://postimg.org/image/f1kxxqprl/
Sticking to cyclists and distance in metres and time in seconds, the cyclist accelerates for 17s up to 4.7m/s then decelerates for a hill for 6s. After the brow of the hill bend they accelerate for 12s up to 1.7m/s. As the slope steepens they accelerate for 5s up to 18.3m/s which is about 40mph.
NB, the units on the y and x axis should match up. If the x axis is in seconds then the y axis should be in (distance measure)/s.
Note that since the distance from the origin of travel increases from 0 to 60m for the cyclist and then decreases back to 0, they must smoothly turn and go back down to the origin, or continue on down a symmetric hill to a symmetric origin. In the first case the route down is exactly the same as the route up in reality. In the second case the route down is the same in its vertical pattern as the route up, but the horizontal pattern is the same in shape but is in a different direction on the compass.
If the distance is in metres and the time minutes the object could be something taken straight up in a hot air balloon (accelerates as more fuel is pumped into the fuel chamber) and then dropped straight down (with some considerable air resistance).
To make the distance graph more realistic add curves as abrupt changes in speed are not common in practice. Here we could have an S-curve from 0s -17s and a parabola from 17s-40s whose maximum is at 25s.