Question

A biker rides on a straight line such that his displacement from the starting point is given as x = 2t + t2. What is the average velocity (average rate of change of displacement) for between t = 4s and t = 6s?
12 m/s
48 m/s
16 m/s
18 m/s
2 m/s

Answer 1

f(6) - f(4) / ( 6 - 4 )

Answer 2

i got it thank you its 12m/s

Answer 3

the answer is right 12 but the way you got it is wrong. the f(x)=2t+t^2 is a displacemnt and not velocity
so to get the velocity you need to take the drv. of f(x) which would be df/dt=2+2t
df/dt(6)=2+2(6)=14
df/dt(4)=2+2(4)=10
Avg=(dr/dt(6)+df/dt(4))/(6-4)=(14+10)/2=12