If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, it height in feet after t second is given by y = 70 t - 16 t^2. find the average velocity for the time period beginning when t = 2 and lasting
(i) 0.1 seconds

Velocity = displacement/ time = y2-y1/t2-t1 
y= 70t-16t^2 we know t= 2 and initial velocity = 70 
so av= y2 ( value of y when we sub in t=2) which for this case y=70(2)-16(2^2)= 76 
therefor 76-70/2.1-2= 60 a) 
and 76-70/2.01-2= 600 b) 
and 76-70/2.001-2=6000 c) 

a) 76-70/0.1 = 60ft 
b) 76-70/0.01= 600ft 
c) 76-70/0.001= 6000ft

B is the answer because the area between the velocity time graph and the x-axis is the distance traveled. If you look at the velocity time graph from 0-2, seconds you will see an area of 40m. Only choice B matches that at 2 seconds, the distance traveled is 40m. You can go on to confirm the remaining trip matches that of B by calculating the 2-4 seconds of the velocity time graph area between the x-axis and graph as -40m (negative because area is under the x-axis). Again, it matches graph B, showing that the total distance traveled is 0 for the entire trip.