In a school experiment, students timed each other as they went various

 

­imRaj
In this question, what assumption says is that 
D2 - D1 / V2 - V1 is constant (3m ~ 15m) For example, for Hopping (10 - 5)/(3.1 - 2.0) = 5/0.9 is constant 

The average speed for hopping 3 meters would be greater than that for speed walking 3 meters.
=> for Speed walking average speed 3m, we can calculate this way 
(2.4 - V) / (5 - 3) = (2.6 - 2.4)/(10-5) result is 2.32. But, average speed for hopping 3 meters will be less than 2m/s 
So, No

The average speed for speed walking 15 meters would be greater than that for hopping 15 meters.
=> average speed for speed walking 15 meters is 2.8 and verage speed for hopping 15 meters is 4.2  
So, No

The average speed for walking backward 15 meters would be greater than that for walking forward 3 meters.
=> average speed for walking backward 15 meters is 2.0 and average speed for walking forward 3 meters is 1.52 
(5-3)/(1.6 - x) = (10-5)/(1.8-1.6)
So, Yes

Though the most wonderful explanation has already been given by my friend EDDIE98 ,
I would like to put forward mine as well.
Basically the distance is proportional to the average speed.
So if the average speed is 2m/s for 5 meters and 3.1 m/s for 10 meters , the average speed will be 4.2 m/s for 15 meters because the speed is increasing in equal proportion i.e (3.1 - 2) = .9 m/s
Please find the diagram attached.
For calculating speed for 3 meter distance, simply put it in the equation and solve.
It will be = .2 m/s­
Hope the diagrams helped.

­wont the increase in speed in equal proportion be (3.1-2)=1.1m/s which is why, the average speed at 15m is (3.1+1.1)=4.2m/s? Good explanation btw